Answer:
Kc = [H₂CO₃] / [CO₂]
Explanation:
Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.
Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.
Also, each specie must be powered to its reactant coefficient.
For example, for the reaction:
aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)
The equilibrium constant, kc is:
Kc = [D]ⁿ / [B]ᵇ[E]ˣ
You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants
Thus, for the reaction:
CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)
The Kc is:
Kc = [H₂CO₃] / [CO₂]
Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the ions that would be present in solution. Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation. Express your answer as a chemical equation including phases.
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
A sample of neon gas at a pressure of 0.609 atm and a temperature of 25.0 °C, occupies a volume of 19.9 liters. If the gas is compressed at constant temperature to a
volume of 12.7 liters, the pressure of the gas sample will be
atm.
Answer:
The pressure of the gas sample will be 0.954 atm.
Explanation:
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
To determine the change in pressure or volume during a transformation at constant temperature, the following is true:
P1 · V1 = P2 · V2
That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.
In this case:
P1= 0.609 atmV1= 19.9 LP2=?V2= 12.7 LReplacing:
0.609 atm* 19.9 L= P2* 12.7 L
Solving:
[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]
P2= 0.954 atm
The pressure of the gas sample will be 0.954 atm.
cetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.
Answer:
0.60 mol
Explanation:
There is some info missing. I think this is the original question.
Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.
Step 1: Given data
Moles of water required: 1.5 mol
Step 2: Write the balanced equation
C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)
Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water
The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
P2O5 (s) + H2O (l) =H3PO4 (aq)
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.
Answer:
1. The coefficients are: 1, 3, 2
2. From the balanced equation, we obtained the following:
3 moles oxygen, O2 reacted.
2 moles of Hydrogen sulfide, H2S reacted.
2 moles of water were produced.
2 moles of sulphur dioxide, SO2 were produced.
Explanation:
1. Determination of the coefficients of the equation.
This is illustrated below:
P2O5(s) + H2O(l) <==> H3PO4(aq)
There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:
P2O5(s) + H2O(l) <==> 2H3PO4(aq)
There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:
P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)
Now the equation is balanced.
The coefficients are: 1, 3, 2.
2. We'll begin by writing the balanced equation for the reaction. This is given below:
2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)
From the balanced equation above,
3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.
11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KCl solution? Are these results what you would expect? Why?
Answer:
Explanation:
KHT is a salt which ionises in water as follows
KHT ⇄ K⁺ + HT⁻
Solubility product Kw= [ K⁺ ] [ HT⁻ ]
product of concentration of K⁺ and HT⁻ in water
In KCl solution , the solubility product of KHT will be decreased .
In KCl solution , there is already presence of K⁺ ion in the solution . So
in the equation
[ K⁺ ] [ HT⁻ ] = constant
when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .
Carbon-14 has a half-life of 5720 years and this is a fast-order reaction. If a piece of wood has converted 75 % of the carbon-14, then how old is it?
Answer:
11445.8years
Explanation:
Half-life of carbon-14 = 5720 years
First we have to calculate the rate constant, we use the formula :
How many mL of calcium hydroxide are required to neutralize 25.0 mL of 0.50 M
nitric acid?
Answer:
6.5 mL
Explanation:
Step 1: Write the balanced reaction
Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O
Step 2: Calculate the reacting moles of nitric acid
25.0 mL of 0.50 M nitric acid react.
[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]
Step 3: Calculate the reacting moles of calcium hydroxide
The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol
Step 4: Calculate the volume of calcium hydroxide
To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.
[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]
below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (g) ΔH1°= 449.2 kJ R2 O3 (g) + Cl (g) → O2 (g) + ClO (g) ΔH2° = -126 kJ R3 ClO (g) + O (g) → O2 (g) + Cl (g) ΔH3°= -268 kJ
Answer:
ΔH = -338.8kJ
Explanation:
it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).
Using the reactions:
R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ
R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ
R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ
By the sum 2R₂ + 2R₃:
(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)
ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ
Now, this reaction + R₁
2O₃(g) → 3O₂(g)
ΔH = -768kJ + 449.2kJ
ΔH = -338.8kJ