Answer:
the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
Explanation:
Given that :
a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]
b. [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]
c. [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]
Acid strength is the ability of an acid to dissociate into a proton and an anion. Take for instance.
HA ↔ H⁺ + A⁻
The acid strength of the following compounds above are:
[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰
[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴
[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵
As the acid dissociation constant increases the relative acid strength also increases.
From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid
Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.
A voltaic cell is set up as follows: Anode: Zn electrode in a solution of 0.050 M Zn(NO 3 ) 2 Cathode: Pt electrode with 0.500 atm H 2 (g) in 0.010 M HNO 3 a) Write the overall balanced cell reaction.
Answer:
[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]
Explanation:
Given that,
Anode : Zn electrode in a solution of 0.050 M Zn(NO₃)₂
Cathode : Pt electrode with 0.500 atm H₂(g) in 0.010 M HNO₃
Anode :
[tex]Zn(s)\Rightarrow Zn^{2+}(aq) + 2e^{-}[/tex]
Cathode :
[tex]2H+(aq)+2e^{-}\Rightarrow H_{2}(g)[/tex]
We need to write the overall balanced cell reaction
Using anode and cathode
[tex]Zn(s) + 2H+(aq)\Rightarrow Zn^{2+}(aq) + H_{2}(g)[/tex]
Hence, This is required answer.
Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(g) 2) H 2O(l) → H 2O(s) 3) Br 2(l) → Br 2(g) 4) H 2O 2(l) → H 2O(l) + 1/ 2O 2(g)
Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]
3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.
2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]
1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.
3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]
1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.
4) [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]
1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.
What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
3. Write the following isotope in hyphenated form (e.g., "carbon-14”): Kr
a. Krypton-109
b. Krypton -37
c. Krypton -36
d. Krypton -73
Answer:
Krypton -73
Explanation:
There are 33 known isotopes of krypton (36Kr) with atomic mass numbers from 69 through 101.
Good luck!
Answer:
D. Krypton-73
Explanation:
An isotope of an element has the same atomic number and the same number of protons but a different number of neutrons and a different atomic weight. Krypton is the 36th element on the periodic table. It has 36 protons and 48 neutrons. Krypton-73 is one of 33 known isotopes of Krypton and is the only one that actually exists from the list of choices.
Hope that helps.
The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a ________ group.
a. carbamate group
b. carbonate
c. carboxlate
d. carboxyl
Answer:
d. carboxyl
Explanation:
The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.
For example:
Formic acid or Methanoic acid (H-COOH) Butanoic acid (C3H7-COOH)Hence, the correct option is "d. carboxyl ".
Assume that 33.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.How many moles of have been added at the equivalence point?n = ? mol
Plz help!!!! Solve this by using factor labeling
Answer:
the answer is 2,000 nickels.
Explanation:
we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.
A baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?
Answer:
I hope it works
Explanation:
As we know that
w=m*g
given m=0.145 , g=9.8
hence we get
w= (9.8)*(0.145)
w=1.421 m/sec 2
if its help-full thank hit the stars and brain-list it thank you
Consider the reaction for the dissolution of solid magnesium hydroxide.
Mg(OH)2(s)g2 (a) +2OH (ag)
If the concentration of hydroxide ion in a saturated solution of magnesium hydroxide is 2.24 x 104 M.
what is the molar solubility of magnesium hydroxide? Report your answer in scientific notation with three significant figures.
Answer:
Molar solubility is 1.12x10⁻⁴M
Explanation:
The dissolution of magnesium hydroxide is:
Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻
The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:
Mg(OH)₂(s) ⇄ X + 2X
Where X is solubility.
If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:
2X = 2.24x10⁻⁴M
X = 2.24x10⁻⁴M/2
X =1.12x10⁻⁴M
Molar solubility is 1.12x10⁻⁴M
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K sp (MgF 2) = 6.9 × 10 –9]
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Draw the major condensation product obtained by treatment of ethyl 3-methylbutanoate with sodium ethoxide in ethanol.
Answer:
ethyl 3-ethoxy-3-hydroxy-2-isopropyl-5-methyl hexanoate
Explanation:
In this case, we have a very strong base (sodium ethoxide). Therefore, this compound will remove a hydrogen from ethyl 3-methyl butanoate generating a carbanion.
This carbanion, can attack another ethyl 3-methyl butanoate molecule on the carbonyl group generating a new C-C bond and producing a negative charge in the oxygen.
Then the ethanol can protonate the molecule generating an "OH" group and the ethoxide.
See figure 1
I hope it helps!
which law states that the pressure and absolue tempeture of a fixed quantity of gas are directly proportional under constant volume conditions?
Answer:
Gay lussacs law
Explanation:
According to the following reaction, how many moles of ammonia will be formed upon the complete reaction of 31.2 grams of nitrogen gas with excess hydrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
4.46 mol of NH3
Explanation:
The equation of he reaction is given as;
2N + 3H2 --> 2NH3
From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.
Mass of Nitrogen = 31.2g
Molar mass of Nitrogen = 14g/mol
Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol
Since 1 mol of N = 2 mol of NH3;
2.23 mol of N2 would produce x
x = 2.23 * 2 = 4.46 mol of NH3
The ph of the test tube can be calculated by knowing the concentration of hydroxide ions in solution. the ph = 14 + log 10 oh- for example the 0.1m stock of naoh has a ph = 14+ log 10 0.1= 13. using the dilution 5 ml 0.1 naoh 5ml h20 what is the ph of tube 1.
Answer:
Volume of the calcium hydroxide solution used is 0.0235 mL.
Explanation:
Moles of KHP =
According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;
of calcium hydroxide
Molarity of the calcium hydroxide solution = 0.703 M
Volume of calcium hydroxide solution = V
Volume of the calcium hydroxide solution used is 0.0235 mL.
write the balanced nuclear equation for the radioactive decay of radium-226 to give radon-222, and determine the type of decay
Answer:
226Ra88→222Rn86+4He2
Explanation:
An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.
During α-decay(alpha decay), an atomic nucleus emits an alpha particle.
1. For the following reaction, 4.86 g of magnesium nitride are mixed with excess water. The reaction yields 7.18 g of magnesium hydroxide.
magnesium nitride(s) + water(1) –> magnesium hydroxide (aq) + ammonia (aq)
What is the ideal yield of magnesium hydroxide?
What is the percent yield for this reaction?
2. For the following reaction, 6.41 g of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 26.2 g of ammonia.
nitrogen(g) + hydrogen(g) –> ammonia(g)
What is the ideal yield of ammonia?
What is the percent yield for this reaction?
3. For the following reaction, 3.79 g of water are mixed with excess chlorine gas. The reaction yields 8.70 g of hydrochloric acid.
chlorine(g) + water(1) –> hydrochloric acid(aq) + chloric acid (HCIO3)(aq)
What is the ideal yield of hydrochloric acid?
What is the percent yield for this reaction?
Answer:
See explanation
Explanation:
1)
Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)
Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles
1 mole of magnesium nitride yields 3 moles of magnesium hydroxide
0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide
Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g
Percent yield= actual yield/ theoretical yield × 100
Percent yield= 7.18/8.398 × 100/1 = 85.5%
2)
N2(g) + 3H2(g) -------> 2NH3(g)
Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.
From the balanced reaction equation;
3 moles of hydrogen gas yields 2 moles of ammonia
3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia
Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g
Percent yield = actual yield/ theoretical yield ×100
Percent yield = 26.2/36.3 ×100
Percent yield = 72.2%
3)
3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)
Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles
Since
3 mole of water yields 5 mole of HCl
0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl
Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g
Percent yield = actual yield/ theoretical yield × 100/1
Percent yield = 8.70/12.775 ×100
Percent yield = 68.1%
What is the oxidizing agent in the redox reaction represented by the following cell notation? Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
Answer:
Silver.
Explanation:
Hello,
In this case, for the redox reaction:
[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]
We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
Best regards.
The oxidizing agent in the redox reaction represented by the following cell notation is Silver.
Calculation of the oxidizing agent:The redox reaction is
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand, the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
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If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
A qualitative researcher may select a ______ case that is unusually rich in information pertaining to the research question.
A) critical
B) typical
C) deviant
D) rare
Answer:
D. Rare
Explanation:
Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.
A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.
This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.
For a spontaneous process, which of the following MUST be true?
A. TΔS>ΔH
B. ΔG>0
C. ΔSuniv>0
D. ΔSsys>ΔSsurr
Answer:
C; ΔSuniv>0
Explanation:
In this question, we want to select which of the options must be true.
What we should understand is that for a process to be spontaneous, the change in entropy must be greater than 0 i.e the change in entropy must be positive.
Looking at the options we have; option C is the correct answer.
Option B looks correct but it is wrong. This is because if change in universal entropy is greater than zero, then change in Gibbs free energy must be less than zero for spontaneity to occur
It can be deduced that for a spontaneous process, B. ΔG>0.
What is a spontaneous process?It should be noted that a spontaneous process simply means a process that occurs without input of matter or electrical energy.
In this case, for a spontaneous process, it's true that ΔG>0, it should be noted that a spontaneous process related to the second law in thermodynamics. This is characterized by an increase in entropy.
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In the experiment students will create solutions with different ratios of ethanol and water. What is the mole fraction of ethanol when 10.00 mL of pure ethanol is combined with 2.00 mL of water
Answer:
[tex]x_{et}=0.6068[/tex]
Explanation:
Hello,
In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:
[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]
Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)
[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]
Therefore, the mole fraction turns out:
[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]
Best regards.
Place the following compounds in order of increasing strength of intermolecular forces. CH4 CH3CH2CH3 CH3CH3 Group of answer choices CH3CH2CH3 < CH4 < CH3CH3 CH4 < CH3CH3 < CH3CH2CH3 CH3CH2CH3 < CH3CH3 < CH4 CH4 < CH3CH2CH3 < CH3CH3 CH3CH3 < CH4 < CH3CH2CH3
Answer:
CH4 < CH3CH3 < CH3CH2CH3
Explanation:
Alkanes are saturated aliphatic hydrocarbons that undergoes intermolecular Van der waals forces. Van der waals forces are the attractive forces which make it possible for non-polar molecules to form liquids and solids.
Van der waals force are described as intermolecular forces arising from induced fluctuating dipoles in atoms and molecules brought about by movement of electrons around the atomic nucleus.
An example of the Van der waals force is the london dispersion force that occurs in the alkane family. It is the weakest of all electrical forces that act between atoms and molecules, These forces are responsible for liquefaction or solidification of non-polar substances at low temperature.
The strength of the intermolecular forces is based on the number of electrons surrounding the molecule and the surface area of the molecule. SO, in alkanes, the longer the carbon chain, the more stronger the intermolecular forces.
how can you prevent frequent landslides from heavy rains
Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.
Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud
B. Plant vegetation on the slops that expreience landslides.
E. Reduce the sloped where landslides occur.
Hope this helped, please mark brainiest!
How many liters of CH₃OH gas are formed when 3.20 L of H₂ gas are completely reacted at STP according to the following chemical reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP.CO(g)+ H2(g) → CH3OH
Answer:
The correct answer is 1.60 Liters.
Explanation:
The given reaction:
CO (g) + H₂(g) ⇔ CH₃OH (g)
Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.
It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.
As 22.4 L at standard temperature and pressure is equivalent to 1 mole.
Therefore, 1 L at STP will be, 1/22.4 mole
Now 3.20 L at STP will be,
= 1/22.4 × 3.20
= 0.1428 mole
And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH
Now, 0.1428 mole of H₂ will give,
= 0.1428/2 = 0.071 mole of CH₃OH
= 0.071 × 22.4 = 1.60 L
The volume, in liters, of CH₃OH gas formed is 1.60 L
From the question,
We are to determine the volume of CH₃OH formed
The given chemical equation for the reaction is
CO(g)+ H₂(g) → CH₃OH
The balanced chemical equation for the reaction is
CO(g)+ 2H₂(g) → CH₃OH
This means
1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH
Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
x mole of the H₂ gas will have a volume of 3.20 L at STP
x = [tex]\frac{3.20 \times 1}{22.4}[/tex]
x = 0.142857 mole
∴ The number of mole of H₂ present is 0.142857 mole
Since
2 moles of H₂ reacts to produce 1 mole of CH₃OH
Then,
0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH
[tex]\frac{0.142857}{2} = 0.0714285[/tex]
∴ The number of moles of CH₃OH produced = 0.0714285 mole
Now, for the volume of CH₃OH formed
Since
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285 at STP
22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L
Hence, the volume of CH₃OH gas formed is 1.60 L
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In which ONE of the following compounds would the bonding be expected to have the highest percentage of ionic character? A) LiBr B) CsCl C) BaBr2 D) NaCl E) KI
Answer:
B) CsCl
Explanation:
The ionic character is formed between two kinds of atoms having a large electronegativity differences e.g metals (like those in groups IA and IIA) and nonmetals (like those in groups VIA and VIIA). The formation of an ionic character involves a transfer of electrons from the less electronegative atom(metal) to the more electronegative atom (non-metal) such that the two kinds of atoms now have completely filled outer shell like the noble gases.
In CsCl, electrons are being transferred from Cs⁺ to Cl⁻ . As a result of this transfer , the atom of the metal becomes positively charged (cation) while that of the non-metal becomes negatively charged (anion).
The highest percentage of ionic character will occur as a result of smaller negatively charged (anion) and larger positively charged (cation). From the options given, CsCl have the highest percentage of ionic character.
clacium hydroxide is slightly soluable in water about 1 gram will dissolve in 1 liter what are the spectator ions in the reaction ions in the reaction of such a dilute solution of calcium hydroxide with hydrochloric acid
Answer:
Ca²⁺ and Cl⁻
Explanation:
In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.
In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:
Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻
The ions that react are H⁺ and OH⁻ (Acid and base producing water)
And the ions that are not reacting, spectator ions, are:
Ca²⁺ and Cl⁻Given that H2(g)+F2(g)⟶2HF(g)ΔH∘rxn=−546.6 kJ 2H2(g)+O2(g)⟶2H2O(l)ΔH∘rxn=−571.6 kJ calculate the value of ΔH∘rxn for
Answer:
ΔH∘rxn of the reaction is -521.6kJ
Explanation:
Complete question: "Calculate the value of ΔH°rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)"
You can find the ΔH of a reaction by the algebraic sum of similar reactions (Hess's law) as follows:
(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔH∘rxn=−546.6 kJ
(2) 2H₂(g)+O₂(g)⟶2H₂O(l) ΔH∘rxn=−571.6 kJ
Subtracting 2ₓ(1) - (2)
2ₓ(1) 2H₂(g) + 2F₂(g) ⟶ 4HF(g) ΔH∘rxn=2ₓ−546.6 kJ = -1093.2kJ
-(2) 2H₂O(l) ⟶ 2H₂(g)+O₂(g) ΔH∘rxn=- (-571.6 kJ) = 571.6kJ
2ₓ(1) - (2) 2F₂(g)+ 2H₂O(l) ⟶ 4HF(g) + O₂(g)
H₂(g) are cancelled because are the same in products and reactants
ΔH∘rxn = -1093.2kJ + 571.6kJ
ΔH∘rxn = -521.6
ΔH∘rxn of the reaction is -521.6kJThe half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed? Report your answer to 1 decimal place.
Answer:
12.5g
Explanation:
Half life = 2.4 Minutes.
The half life of a compound is the time it takes to decay to half of it's original concentration or mass.
Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)
Initial mass = 100g
First half life;
100g --> 50g
Second half life;
50g --> 25g
Third half life;
25g --> 12.5g
The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2.4 mins
Time (t) = 7.2 mins
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]
n = 3Thus, 3 half-lives has elapsed.
Finally, we shall the amount remaining. This can be obtained as follow:Original amount (N₀) = 100 g
Number of half-lives (n) = 3
Amount remaining (N) =?[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]
N = 12.5 gThus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g
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Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane is burned? Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer. Equation: C3H8 + 5O2 ->>>>>>> 3CO2 + 4H2O
Answer:
Explanation:
Number of moles of propane:
=Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
= 44 * 6.8181
= 299.9964 grams
A number of moles of propane:
Mass in grams ÷ Relative molecular Mass
= 100/((12*3) + (1*8))
= 100 ÷ 44
= 2.2727
Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)
Number of moles of CO2 = 3*2.2727 = 6.8181
Mass in grams = Relative molecular Mass * Number of moles
=44 * 6.8181
= 299.9964 grams
What is carbon dioxide useful for?Carbon dioxide is used as a refrigerant, in fireplace extinguishers, for inflating lifestyles rafts and life jackets, blasting coal, foaming rubber and plastics, selling the increased vegetation in greenhouses, and immobilizing animals earlier than slaughter, and in carbonated liquids.
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The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.1 grams of 238U, but very little 235U. (a) What is the activity of the plate (in Ci)
Answer:
The correct answer is 1.68 × 10⁻⁵ Ci
Explanation:
The activity of the uranium is determined by using the formula,
R = 0.693 N/t1/2 -------------- (i)
The number of atoms is, N = nNA
Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,
N = (m/M) NA
The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,
N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³
The half-life of 238U from year to second is,
t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s
Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,
R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s
= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)
= 1.68 × 10⁻⁵ Ci
Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci