Answer:
let the people who dont know how to fight on the life boat and the fighters can stay back and try to keep the sharks away.
Name four types of salts
Answer:
Any ionic molecule formed of a base and an acid, which dissolves in water to produce ions is known as a salt. The four common types of salts are:
1. NaCl or sodium chloride is the most common kind of salt known. It is also known as table salt.
2. K2Cr2O7 or potassium dichromate refers to an orange-colored salt formed of chromium, potassium, and oxygen. It is toxic to humans and is also an oxidizer, which is a fire hazard.
3. CaCl2 or calcium chloride looks like table salt due to its white color. It is broadly used to withdraw ice from roads. It is hygroscopic.
4. NaHSO4 or sodium bisulfate produces from hydrogen, sodium, oxygen, and sulfur. It is also known as dry acid. It has commercial applications like reducing the pH of swimming pools and spas and others.
Why don't siblings look exactly alike
Answer:
Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.
Why are cells important to an organisms survival
Answer:
Cells are the basic structures of all living organisms. Cells provide structure for the body, take in nutrients from food and carry out important functions. ... These organelles carry out tasks such as making proteins?, processing chemicals and generating energy for the cell
Answer: I absolutely love this question! Biology is so interesting, so I always love to answer the curiosity of others regarding biology, such as that!
Cells are simply the basic structures of all organisms, that are living, of course! Cells provide structure for the body, and they also take in nutrients that your body needs from food and they carry out important functions. These organelles carry out tasks such as making proteins, processing chemicals, and generating energy for the cell. Isn’t that cool?
Hope this helped! <3
Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
Answer:
Answers are in the explanation
Explanation:
It is possible to obtain K of equilibrium of related reactions knowing the laws:
A + B ⇄ C K₁
C ⇄ A + B K = 1 /K₁
The inverse reaction has the inverse K equilibrium
2A + 2B ⇄ 2C K = K₁²
The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients
For the reaction:
2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K
1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)
This is the inverse reaction but also the coefficients are dividing in the half, that means:
[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]
2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)
Here,the only change is the coefficients are the half of the original reaction:
[tex]K_2 = K^{1/2}[/tex]
3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)
This is the inverse reaction. Thus, you have the inverse K of equilibrium:
[tex]K_3 = \frac{1}{K}[/tex]
A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries 20 and 9; column 2 is number of neutrons, with entries 20 and D; Column 3 is atomic number, with entries A and E; Column 4 is Mass Number, with entries B and 19, and Column 5 is Element (symbol) with entries C and F. Using the periodic table, complete the table to describe each atom. Type in your answers
Answer:
A ⇒ 20
B ⇒ 40
C ⇒ Ca
D ⇒ 10
E ⇒ 9
F ⇒ F
Explanation:
edge 2021
Answer:
the person above is correct
Explanation:
For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
Answer:
[tex]m_{CaCO_3}=0.179gCaCO_3[/tex]
Explanation:
Hello,
In this case, since the undergoing chemical reaction is:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:
[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2[/tex]
Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:
[tex]m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3[/tex]
Regards.
The mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g
From the question,
We are to determine the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP.
First, we will determine the number of mole of CO₂ required to be produced
From the formula
PV = nRT
Where
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
and T is the temperature
Then, we can write that
[tex]n = \frac{PV}{RT}[/tex]
From the question,
V = 40.0 mL = 0.04 L
At STP
P = 1 atm
T = 273.15 K
and
R = 0.08206 L atm mol⁻¹ K⁻¹
Putting the parameters into the formula, we get
[tex]n = \frac{1 \times 0.04}{0.08206 \times 273.15}[/tex]
∴ n = 0.0017845 mole
Now, we will write the balanced chemical equation for the decomposition of CaCO₃
CaCO₃ → CaO + CO₂
This means,
1 mole of CaCO₃ will decompose to produce 1 mole of CO₂
Since 0.0017845 mole of CO₂ is to be produced,
Then,
0.0017845 mole of CaCO₃ would be required
Now, for the mass of CaCO₃ required,
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CaCO₃ = 100.0869 g/mol
∴ Mass of CaCO₃ required = 0.0017845 × 100.0869
Mass of CaCO₃ required = 0.178605 g
Mass of CaCO₃ required ≅ 0.179 g
Hence, the mass of CaCO₃ required to produce 40.0 mL of CO₂ at STP is 0.179 g
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A laboratory assistant needs to prepare 217 mL of 0.246 M solution. How many grams of calcium chloride will she need
Answer:
5.92 g
Explanation:
Convert milliliters to liters.
217 mL = 0.217 L
Since molarity (M) is moles per liter(mol/L), multiply the molarity by the volume to find out how many moles you will need.
0.217 L × 0.246 M = 0.05338 mol
Now, convert the moles to grams using the molar mass. The molar mass of calcium chloride is 110.98 g/mol.
0.05338 mol × 110.98 g/mol = 5.924 g ≈ 5.92 g
You will need 5.92 g of calcium chloride.
Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average
Answer:
The correct answer is 2.2 mL.
Explanation:
Given:
Average: 2.9 mL
SD: 0.71 mL
We can define a 1 SD range in which the value of volume (in mL) will be comprised:
Volume (mL) = Average ± SD = (2.9 ± 0.7) mL
Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL
Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL
Thus, the minimum value within a 1 SD range of the average is 2.2 mL
The minimum value within 1 SD is 2.19 mL
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 2.9 mL, σ = 0.71 mL; hence:
The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)
Therefore the minimum value within 1 SD is 2.19 mL
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Given the following reaction and data, A + B → Products
Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640
Required:
a. What is the rate law of the reaction?
b. What is the rate constant?
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
Where X is reaction coordinate
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
[Cu²⁺] = 2.01x10⁻²⁶Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
A solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium? Express the pH numerically.
Answer:
[tex]pH=2.28[/tex]
Explanation:
Hello,
In this case, for the acid dissociation of formic acid (HCOOH) we have:
[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]
Whose equilibrium expression is:
[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
That in terms of the reaction extent is:
[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]
Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:
[tex]x=0.00528M[/tex]
[tex][H^+]=0.00528M[/tex]
Then, as the pH is computed as:
[tex]pH=-log([H^+])[/tex]
The pH turns out:
[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]
Regards.
Which ONE of these cations has the same number of unpaired electrons as Fe2+ ? A) Ni2+ B) Fe3+ C) Cr2+ D) Mn2+ E) Co2+
Answer:
Explanation:
Fe2+ Has 4 unpaired electrons.
By method of elimination;
Option A: Ni2+ has two unpaired electrons. so this option is wrong.
Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.
Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.
Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.
Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?
Explanation:
To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;
Step 1: Divide the percentage composition by the atomic mass
Sulphur = 31.42 / 32 = 0.9819
Oxygen = 31.35 / 16 = 1.9594
Flourine = 37.23 / 19 = 1.9595
Step 2: Divide by the lowest number
Sulphur = 0.9819 / 0.9819 = 1
Oxygen = 1.9594 / 0.9819 ≈ 2
Flourine = 1.9595 / 0.9819 ≈ 2
This means the ratio of the elements is 1 : 2: 2
The empirical formular (simplest formular of a compound) of the compound is;
SO₂F₂
To obtain the molecular formular (Actual formular of a compound);
(SO₂F₂)n = 102.1
Inserting the atomic masses and solving for n;
(102)n = 102.1
n ≈ 1
The molecular formular is; (SO₂F₂)₁ = SO₂F₂
Given that the Ksp value for Ca3(PO4)2 is 8.6×10−19, if the concentration of Ca2+ in solution is 4.9×10−5 M, the concentration of PO3−4 must exceed _____ to generate a precipitate.
Answer:
.0027 M
Explanation:
We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)
Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2
[PO3−4]=0.0027 M
One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
If a sample of C-14 initially contains 1.6 mmol of C-14, how many millimoles will be left after 2250 years
Answer: 1.2 millimoles will be left after 2250 years
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]
b) Amount left after 2250 years
[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]
[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]
[tex]\log\frac{1.6}{a-x}=0.117[/tex]
[tex]\frac{1.6}{a-x}=1.31[/tex]
[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]
Thus 1.2 millimoles will be left after 2250 years
Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation:
To find the pH of a solution of NaNO2, one would have to construct an ICE chart using:
a. the Kb of NO−2 to find the hydroxide concentration.
b. the Kb of HNO2 to find the hydronium concentration.
c. the Kb of NO-2, to find the hydronium concentration.
d. the Kb of HNO2, to find the hydroxide concentration.
Answer:
a. the Kb of NO₂⁻ to find the hydroxide concentration.
Explanation:
When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.
NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)
Na⁺ comes from NaOH (strong base) so it doesn't react with water.
NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to find the hydroxide concentration.
A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
Read the article. Use your understanding to answer the questions that follow. What type of source is this article? primary or secondary and how do you know
Answer: C
Explanation:
The article was sourced from the Oak National Laboratory
Which reasons did you include in your response? Check all of the boxes that apply.
1. The article does not present original research.
and
3. The article has references to primary sources.
Answer:
C
Explanation:
Which reasons did you include in your response? Check all of the boxes that apply.
The article does not present original research.
The article summarizes other research.
The article has references to primary sources.
The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?
Answer:
2.8x10⁻¹⁵ M.
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]
In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:
[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]
Thus, solving for [tex]x[/tex] we have:
[tex]x=2.8x10^{-15}M[/tex]
Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.
Regards.
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
21. What are the two main ways of working with clay?
Answer:
Diferentes tipos de arcilla
ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
Coil method and the slab method.
Explanation:
To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At which electrode does oxygen form? options: A) Both the anode and the cathode B) Cathode C) Neither electrode D) Anode
Answer:
im pretty sure its the anode
Explanation:
To solve such, we must know the concept of electrolysis reaction. The correct option is option D among all given options. At anode electrode oxygen forms.
What is chemical reaction?Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.
Electrolysis is the process of passing an electric current through a material to cause a chemical change. A chemical change occurs when a material loses or acquires the electron. To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At anode electrode oxygen forms.
Therefore, the correct option is option D among all given options. At anode electrode oxygen forms.
Learn more about the chemical reactions, here:
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the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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When salt is added to water, all of the following happens except? A. The salt breaks into positive chlorine ions and negative sodium icons B. the positive part of the water molecule is attracted to the negative ions C. The negative part of the water molecule is attracted to the positive ions D. The water molecules surround the dissociated ions
Answer:
The salt breaks into positive chlorine ions and negative sodium icons
Explanation:
The question requested for the wrong option in the list. If we look at the option selected, we will notice that sodium ions are positively charged ions since sodium is a metal. Metals produce cations (positive ions) because they loose electrons. Therefore, a sodium ion can never be negatively charged.
Similarly, chlorine is a highly electronegative nonmetal. It gains electrons in an ionic bond. Hence chlorine ions can not be positive.