Answer:
See explanation
Explanation:
Many organic compounds have low melting points. This is due to the fact that many of these compounds are non polar.
However, compound X is slightly polar but still has a melting point which is far less than that of sand composed of a high melting point inorganic material.
Since sand has a much higher melting point compared to compound X, the researcher need not be worried that sand was spilled into his beaker.
What is the unite of Kc ?
Answer:
There is no unit for Kc
Explanation:
The rete at which someone or something moves or travels
Speed.
Speed is the rate at which someone or something moves or travels.
When an electron is displaced in a semi conductor, the hole that’s left behind is?
Explanation:
holes are left due to space left out the migrating electrons. they are positive charge carriers. if the electron moves in one direction, the holes appear to move in the opposite direction.
Determine the empirical formula of a compound having the following percent composition by mass: K: 24.74%; Mn: 34.76%; O: 40.50%
Answer: The empirical formula of the compound is [tex]KMnO_4[/tex]
Explanation:
The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Let the mass of the compound be 100 g
Given values:
% of K = 24.7%
% of Mn = 34.76%
% of O = 40.50%
Mass of K = 24.7 g
Mass of Mn = 34.76 g
Mass of O = 40.50 g
To calculate the empirical formula of a compound, few steps need to be followed:
Step 1: Calculating the number of moles of each elementWe know:
Molar mass of K = 39.10 g/mol
Molar mass of Mn = 54.94 g/mol
Molar mass of O = 16 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol[/tex]
[tex]\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol[/tex]
[tex]\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol[/tex]
Step 2: Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles[tex]\text{Mole fraction of K}=\frac{0.632}{0.632}=1[/tex]
[tex]\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1[/tex]
[tex]\text{Mole fraction of O}=\frac{2.53}{0.632}=4[/tex]
Step 3: Writing the mole fraction as the subscripts of each of the elementThe empirical formula of the compound becomes [tex]K_1Mn_1O_4=KMnO_4[/tex]
Hence, the empirical formula of the compound is [tex]KMnO_4[/tex]
The empirical formula of the compound is KMnO₄
Data obtained from the question Potassium (K) = 24.74%Manganese (Mn) = 34.76%Oxygen (O) = 40.50%Empirical formula =? How to determine the empirical formulaDivide by their molar mass
K = 24.74 / 39 = 0.634
Mn = 34.76 / 55 = 0.632
O = 40.50 / 16 = 2.531
Divide by the smallest
K = 0.634 / 0.632 = 1
Mn = 0.632 / 0.632 = 1
O = 2.531 / 0.632 = 4
Therefore, the empirical formula of the compound is KMnO₄
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What makes up the central nervous system?
a.
motor neurons.
b.
interneurons.
c.
brain and spinal cord.
d.
liver and pancreas.
Answer:
c.
brain and spinal cord.
Explanation:
Note: Please show all work and calculation setups to get full credit. T. he following may be used on this assignment: specific heat of (water=4.184 J/g oC; ice=2.03 J/g oC; steam=1.99 184 J/g oC); heat of fusion of water=80. cal/g; heat of vaporization=540 cal/g; 1cal=4.184J.
Calculate the energy required (in J) to convert 25 g of ice at -15 oC to water at 75 oC.
Answer:
1.7 × 10⁴ J
Explanation:
Step 1: Calculate the heat required to raise the temperature of ice from -15 °C to 0°C
We will use the following expression.
Q₁ = c(ice) × m × ΔT
Q₁ = 2.03 J/g.°C × 25 g × [0°C - (-15°C)] = 7.6 × 10² J
Step 2: Calculate the heat required to melt 25 g of ice
We will use the following expression.
Q₂ = C(fusion) × m
Q₂ = 80. cal/g × 25 g × 4.184 J/1 cal = 8.4 × 10³ J
Step 3: Calculate the heat required to raise the temperature of water from 0°C to 75 °C
We will use the following expression.
Q₃ = c(water) × m × ΔT
Q₃ = 4.184 J/g.°C × 25 g × (75°C - 0°C) = 7.8 × 10³ J
Step 4: Calculate the total heat required
Q = Q₁ + Q₂ + Q₃
Q = 7.6 × 10² J + 8.4 × 10³ J + 7.8 × 10³ J = 1.7 × 10⁴ J
. boiling 21.1 g of water at 100°C
How many joules are required to boil 21.1 g of water at 100°C?
Awnser:47,700 J
ASAP!!!!! The data table below shows a person’s heart rate measured in beats per minute (bpm) at five different times in the beginning of a day. What causes the change in heart rate over time?
1. tired
2. increased activity
3. passing of time
4. forgetting to eat
Give two ways to make salt conduct.
Answer:
I think u willl get the answer fast in......
Answer:
this is because when a salt dissolves, its dissociated ions can move freely in solution, allowing a charge to flow. The resulting solution will conduct electricity because it contains ions.
Explanation:
How many moles of AICI3 are produced?
Answer:
please correct it, the question is incomplete
A particle that travels around the nuceleus of an atom in orbitals is called?
I need help and don’t understand, where does each chemical reaction go?
Answer:For the 1st box it starts with 250 and for the 2nd box it starts with CO(2).
Explanation:
How many grams of potassium chloride will be needed to produce
829 grams of zinc chloride?
Answer:
[tex]2KCl + Zn {}^{2 + } → 2K {}^{ + } + ZnCl _{2} \\ molecular \: mass \: of \: zinc \: chloride = 65 + (35.5 \times 2) = 136 \: g \\ molecular \: mass \: of \: potassium \: chloride = 39 + 35.5 = 74.5 \: g
Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2 (g) Identify the excess reactant, the limiting reactant, and the product(s) using the molecular art. (Black spheres are carbon, blue spheres are nitrogen, and red spheres are oxygen.)
The image is not given in the question, it is attached below:
Answer: The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.
Explanation:
In the given image:
Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms
The combination of 1 black and 2 red spheres will represent carbon dioxide [tex](CO_2)[/tex] compound
The combination of 2 blue spheres will represent nitrogen molecule [tex](N_2)[/tex]
The combination of 1 blue and 1 red sphere will represent nitrogen monoxide [tex](NO)[/tex] compound
The combination of 1 black and 1 red sphere will represent nitrogen monoxide [tex](NO)[/tex] compound
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
We are given:
Given moles of NO = 6 moles
Given moles of CO = 4 moles
For the given chemical equation:
[tex]2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)[/tex]
By stoichiometry of the reaction:
If 2 moles of CO reacts with 2 moles of NO
So, 4 moles of CO will react with = [tex]\frac{2}{2}\times 4=4mol[/tex] of NO
As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, CO is considered a limiting reagent because it limits the formation of the product.
Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.
suggest a suitable electrolyte that could be used if the iron ring is to be plated with silver
Nitrogen gas and chlorine gas will react to form nitrogen monochloride gas. 5 moles of nitrogen and 10 moles of chlorine are mixed in a previously evacuated 2 L flask at 500 K. After equilibrium has been established, there are 3 moles of NCl (g) present. What is the equilibrium constant (Kc) for this reaction at 500 K
Answer:
0.30
Explanation:
Step 1: Calculate the known initial and equilibrium molar concentrations
[N₂]i = 5 mol/2 L = 2.5 M
[Cl₂]i = 10 mol/2 L = 5 M
[NCl]e = 3 mol/2 L = 1.5 M
Step 2: Make an ICE chart
N₂(g) + Cl₂(g) ⇄ 2 NCl(g)
I 2.5 5 0
C -x -x +2x
E 2.5-x 5-x 2x
Step 3: Find the value of x
We know that [NCl]e = 2x = 1.5 ⇒ x = 0.75 M
Step 4: Calculate the concentrations at equilibrium
[N₂]e = 2.5-x = 2.5-0.75 = 1.75 M
[Cl₂]e = 5-x = 5-0.75 = 4.25 M
[NCl]e = 1.5 M
Step 5: Calculate the equilibrium constant (Kc)
Kc = [NCl]² / [N₂] × [Cl₂]
Kc = 1.5² / 1.75 × 4.25 = 0.30
Which is most likely to happen during a double-displacement reaction?
O A. Two substances will combine to form one compound.
B. The ions in two compounds will switch places to form two new
compounds.
C. An element will replace another element in a compound.
D. A compound will break apart into two or more substances
I think the Answer is B
In a double-displacement reaction, the cations and anions switch places to produce the product. The ions swap their place to yield new compounds. Thus, option B is correct.
What is a double-displacement reaction?A double-displacement reaction, also called salt metathesis reaction is a process involving the exchange of the bonds and the positive and the negative ions to result in compounds different from the reactant.
The positively charged cation and the negatively charged anion swap their places and produce new products or compounds. It can be of three types namely neutralization, precipitation, and gas formation.
Therefore, the ions (cations and anions) switch their places in a double-displacement reaction.
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They allow atoms to _______ and form ____________ or ____________ substance.
Answer:
They allow atoms to combine and form new substance.
A 1M solution is made of each item in a pair. When would this combination create a buffer? Chose the pairs below that you could use to make a buffered solution.
a. HCI/NaOH
b. HCOOH/ NaHCOO
c. HNO2/H2SO3
d. NaNO3/ HNO3
Answer:
b. HCOOH/ NaHCOO.
Explanation:
A buffer system may be formed in one of two forms:
A weak acid with its conjugate base.A weak base with its conjugate acid.Chose the pairs below that you could use to make a buffered solution.
a. HCI/NaOH. NO. HCl is a strong acid and NaOH is a strong base.
b. HCOOH/ NaHCOO. YES. HCOOH is a weak acid and HCOO⁻ (coming from NaHCOO) is its conjugate base.
c. HNO₂/H₂SO₃. NO. Both are acids and they are unrelated to each other.
d. NaNO₃/ HNO₃. NO. HNO₃ is a strong acid.
What is the total number of atoms in one formula unit of MgSO4.7H2O?
Answer:
11.
O4 means 4 atoms of oxygen. H2O
has one atom of oxygen, so seven "units" of
H2O has 7.
The solubility of compound A in water is 3 g /100ml at 25oC, 11 g/100 ml at 60 oC . If you were to dissolve 20 g of A in 200 ml of water at 60 oC and then slowly cool down the solution to induce crystallization, how many grams of compound A would come out of solution at 25oC?
Answer:
14g of A would come out
Explanation:
The solubility of A at 60°C is 11g/100mL. In 200mL, solubility is:
22g/200mL. That means, the 20g of A will be dissolved completely at 60°C.
Then, at 25°C, the solubility of A is 6g/200g. That means the amount of A that precipitates during the temperature decreasing is:
20g-6g =
14g of A would come out
An 11.5 mL portion of 0.162 M H3PO4 (aq) is to be titrated with 0.229 M NaOH (aq). What volume (in mL) of NaOH will it take to reach the equivalence point
Answer:
8.14 mL of NaOH are required to reach the equivalence point.
Explanation:
Formula for titration is
mmoles of base = mmoles of acid.
Equality is fullfilled at the equivalence point.
M of acid . volume of acid = M of base . volume of base
Let's replace with data given:
11.5 mL . 0.162M = 0.229 M . volume of base
Volume of base = (11.5 mL . 0.162M ) / 0.229M
Volume of base = 8.14 mL
8.14 mL of NaOH are required to reach the equivalence point.
4. How many electrons are in the electron shell closest to the nucleus?
Answer:
2
Explanation:
there are 2 electrons close to the nucleus
Answer:
2
Explanation:
2*n^2, where n is the electron shell.
If we plug 1 in, we will get 2*1, which is 2.
A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a solution made of of alanine () dissolved in of .
The question is incomplete, the complete question is;
A certain liquid X has a normal freezing point of 5.40 °C and a freezing point depression constant K=3.05 ?kg mol solution made of 46.1g of alanine (C3H,NO2) dissolved in 750. g of i. Round your answer to 3 significant digits.
Answer:
3.3 °C
Explanation:
If ∆T= K m I
Where;
∆T = freezing point depression
K = freezing point constant
m= morality of solution
i= Van't Hoff factor
Number of moles of solute = 46.1 g/89.09 g/mol = 0.52 moles
Molality of solution = 0.52 moles /750 ×10-3Kg
= 0.69 m
∆T= 3.05 × 0.69 × 1
∆T= 2.10 °C
Freezing point of solution the solution = freezing point of pure solvent - freezing point depression
= 5.40°C - 2.10 °C
= 3.3 °C
What is the pressure, in atm, exerted by 41.6 g oxygen in a 22.0L container at 30.0C ?
Answer:
1.47 atm
Explanation:
Step 1: Calculate the moles corresponding to 41.6 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
41.6 g × 1 mol/32.00 g = 1.30 mol
Step 2: Convert 30.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 30.0 + 273.15 = 303.2 K
Step 3: Calculate the pressure exerted by the oxygen
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 1.30 mol × (0.0821 atm.L/mol.L) × 303.2 K / 22.0 L = 1.47 atm
What happens to the rate of a reaction as the reaction progresses?
O A. The rate increases as products are formed.
O B. The rate stays constant throughout the reaction.
O C. The rate is constant until it suddenly drops to zero.
O D. The rate decreases as reactants are used up.
Answer:
D. The rate decreases as reactants are used up.
Explanation:
Initially, the rate increases until the reaction is at equilibrium. At equilibrium, the rate is constant.
As the reaction progresses, the rate decreases to zero when reactants are used up ( for irriversible reactions only )
how many grams of NaOH will make 1M of its aqueous solution in 1000cm3
Answer: The mass of NaOH is 40 g
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)
Given values:
Molarity of NaOH = 1 M
Molar mass of NaOH = 40 g/mol
Volume of the solution = [tex]1000cm^3=1000 mL[/tex] (Conversion factor: [tex]1cm^3=1mL[/tex])
Putting values in equation 1, we get:
[tex]1=\frac{\text{Mass of NaOH}\times 1000}{40\times 1000}\\\\\text{Mass of NaOH}=\frac{1\times 40\times 1000}{1000}=40g[/tex]
Hence, the mass of NaOH is 40 g
When the nuclide polonium-218 undergoes alpha decay:
a. The name of the product nuclide is _____.
b. The symbol for the product nuclide is _____.
Write a balanced nuclear equation for the following: The nuclide polonium-218 undergoes alpha emission.
Answer:
a): The name of the product nuclide is lead-214
b): The symbol of the product nuclide is Pb-218
Explanation:
There are three types of decay processes:
Alpha decayBeta decayGamma decayAlpha decay is the decay process that happens when a heavy nucleus decays into a light nucleus with the release of an alpha particle. This alpha particle carries a charge of +2 units and has a mass of 4 units. It is also known as the helium nucleus. The general equation for this decay process is:
[tex]_Z^A\textrm{X} → _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha[/tex]
The nuclear equation for the alpha decay of Po-218 follows:
[tex]_{84}^{218}\textrm{Po}\rightarrow _{82}^{214}\textrm{Pb}+_2^4\alpha[/tex]
Hence, the name of the product nuclide is lead-214 and the symbol is Pb-218.
Make tag question.
neither of them plays cricket,......?
Answer:
neither of them plays cricket,does anyone?
Explanation:
I know you will think I its weird but it is the rule of grammar and 100% correct!
Answer:
neither of them plays cricket, do you?
Explanation:
if you start with .235 g nickel, tell me how much of the compound product will be produced.
Please show work
Answer:
Explanation:
There are many compounds that are produced from Nickel.
Molar mass of Ni = 58.7g
So for 0.235g Ni, there are 0.004M.
By conservation of mass, the compound produced will contain 0.04M of Ni.
Molar mass of nickel=58g/mol
Given mass=0.235g[tex]\\ \sf\longmapsto No\:of\:moles=\dfrac{Given\:Mass}{Molar\:Mass}[/tex]
[tex]\\ \sf\longmapsto \dfrac{0.235}{58}[/tex]
[tex]\\ \sf\longmapsto 0.004mol[/tex]
Using law of conservation of mass the compound will contain 0.04moles of Ni