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kvv77 [185]
3 years ago
7

Balance the following equations. Do not include the states of matter. (a) C + O2 → CO

Chemistry
2 answers:
krek1111 [17]3 years ago
7 0

Answer:

C + O2 → CO2

Explanation:

C + O2 → CO ----------------- (1)

from equ (1) on reactant side, C has 1 mole, O has 2 moles

from equ (1) on product side, C has 1 mole, O has 1 mole

Thus, to balance the equation, O should have 2 moles

C + O2 → CO2

serious [3.7K]3 years ago
6 0

Answer:

C + O2 → CO2

Explanation:

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Which of the following is NOT a chemical reaction?
Natali [406]
Slicing carrotsin a chemical reacționat the reactants, which are on the left side of the equation should become something else which are the products on the right side of the equation!
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If 8cm^3 of H2 reacts with an excess of Cl2, calculate how much of the HCL(g) is produced.
Rom4ik [11]
First, we need the balanced equation: H₂ + Cl₂ ---> 2HCl

since not much information is given, I am assuming we are at STP and that 22.4 Liters= 1 mol

1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.

8 cm³ (1 ml/ 1 cm³)(1 L/ 1000 mL) (1 mol/ 22.4 Liters)= 3.6x10⁻⁴ moles of H₂

2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl

3.6x10⁻⁴ mol H₂ (2 mol HCl/ 1 mol H₂)= 7.1x10⁻⁴ mol HCl

3) lastly, we convert the moles of HCl to grams using the molar mass.

molar mass of HCl= 1.01 + 35.5= 36.51 g/mol

7.1x10⁻⁴ mol HCl (36.51 g/mol)=<span> 0.026 grams HCl</span>
8 0
3 years ago
The value of Ksp for PbCl2 is 1.6 ×10-5. What is the lowest concentration of Cl-(aq) that would be needed to begin precipitation
Dovator [93]

Answer:

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

Explanation:

Concentration of lead nitarte = [Pb(NO_3)_2]=0.010 M

Pb(NO_3)_2(aq)\rightleftharpoons Pb^{2+}(aq)+2NO_{3}^{-}(aq0

1 Mole of lead nirate gives 1 mole of lead ion.

Concentration of lead ion in the solution = 1\times 0.010 M= 0.010 M

Pb(Cl)_2(aq)\rightleftharpoons  Pb^{2+}(aq)+2Cl^{-}(aq0

Concentration of chloride ions = [Cl^-]

The value of K_{sp} for [tex]PbCl_2= 1.6\times 10^{-5}

K_{sp}=[Pb^{2+}][Cl^{-}]^2

1.6\times 10^{-5}=0.010 M\times [Cl^{-}]^2

[Cl^-]=0.04 M

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

3 0
4 years ago
Consider the following reaction where Kc = 1.29×10-2 at 600 K: COCl2 (g) CO (g) + Cl2 (g) A reaction mixture was found to contai
vaieri [72.5K]

Answer:

1. In order to reach equilibrium COCl₂(g) must be consumed.

B. False

2. In order to reach equilibrium Kc must increase.

B. False .

3. In order to reach equilibrium CO must be consumed.

A. True.

4. Qc is greater than Kc.

A. True

5. The reaction is at equilibrium. No further reaction will occur.

B. False.

Explanation:

Based on the reaction:

COCl₂(g) → CO (g) + Cl₂(g)

And Kc is defined as:

Kc = 1.29x10⁻² = [CO] [Cl₂] / [COCl₂]

Molar concentrations of each species are:

[COCl₂] = 0.104 moles of COCl₂ / 1L = 0.104M

[CO] = 4.66×10⁻² moles of CO / 1L = 4.66×10⁻²M

[Cl₂] = 3.76×10⁻² moles of Cl₂ / 1L = 3.76×10⁻²M

Replacing in Kc formula:

4.66×10⁻²M × 3.76×10⁻²M / 0.104M = 1.68x10⁻²

As the concentrations are not in equilibrium, 1.68x10⁻² is defined as the <em>reaction quotient, Qc</em>.

As Qc > Kc, the reaction will shift to the left producing more COCl₂ and consuming CO and Cl₂. Thus

1. In order to reach equilibrium COCl₂(g) must be consumed.

B. False

2. In order to reach equilibrium Kc must increase.

B. False . Kc is a constant that never change.

3. In order to reach equilibrium CO must be consumed.

A. True.

4. Qc is greater than Kc.

A. True

5. The reaction is at equilibrium. No further reaction will occur.

B. False. The reaction is in equilibrium when Qc = Kc

6 0
4 years ago
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