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Maurinko [17]
3 years ago
11

What is the atom inventory for the following equation after it is properly balanced? ____NaOH + ____CuCl2 Imported Asset ____NaC

l + ____Cu(OH)2 Reactants: Na = 1, O = 1, H = 1, Cu = 2, Cl = 2; Products: Na = 1, O = 1, H = 1, Cu = 2, Cl = 2 Reactants: Na = 1, O = 2, H = 2, Cu= 2, Cl= 2; Products: Na = 1, Cl = 1, Cu = 2, O = 2, H = 2 Reactants: Na = 1, O = 1, H = 1, Cu = 1, Cl = 21; Products: Na = 1, O = 1, H = 1, Cu = 1, Cl = 2 Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2; Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2
Chemistry
1 answer:
faltersainse [42]3 years ago
4 0

Answer:

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2;

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

Explanation:

NaOH + CuCL2 —> NaCl + Cu(OH)2

The balanced equation can be achieved by doing the following:

There are 2 oxygen and 2 hydrogen atom on the right side. This is balanced by putting 2 in front of NaOH as shown below:

2NaOH + CuCL2 —> NaCl + Cu(OH)2

This makes Na to be unbalanced. Now to balance Na, put 2 in front of NaCl as illustrated below

2NaOH + CuCL2 —> 2NaCl + Cu(OH)2

Now the equation is balanced.

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

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I think it might be a decomposition.
4 0
3 years ago
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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. what volume of hydrogen chloride
Anna71 [15]
The reaction between methane gas and chlorine gas to form hydrogen chloride and carbon tetrachloride, all in their gaseous form can be expressed through the chemical reaction below.

        CH₄ + 4Cl₂ --> 4HCl + CCl₄

Let us assume that all the involved gases behaves ideally such that each mole of the gas is equal to 22.4 L. 

Through proper dimensional analysis, the volume of the produced hydrogen chloride is calculated,
    V(HCl) = (1.69 mL CH₄)(1 L CH₄/ 1000 mL CH₄)(1 mol CH₄/22.4 L CH₄)(4 mols HCl/1 mol CH₄)(22.4 L HCl/1 mol HCl)(1000 mL/1 L)
    

    V(HCl) = 6.76 mL

<em>ANSWER: 6.76 mL</em>


3 0
3 years ago
How many moles of NaCl are in 75.0 g of NaCl ?
LenaWriter [7]
Number of moles is found by formula n=mass/molar mass, or m/M. the molar mass is found by adding together the atomic masses of Na and Cl (22.99 + 35.45) to give 58.44 g/mol. Since the mass of NaCl is 75.0g, we find the number of moles as follows:
n = 75.0 / 58.44 = 1.28 mol
6 0
2 years ago
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Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =
Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be \frac{-196}{2}\times 1=-98kJ

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol

  • To calculate the heat of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released

n = number of moles = 0.0735 moles

\Delta H_{rxn} = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ

Hence, the amount of heat released is -7.203 kJ

8 0
3 years ago
How many moles is 34.4 g of Oxygen gas (O2)?
jasenka [17]

Answer:

n = 1.075  moles

Explanation:

Given that,

Mass of oxygen = 34.4 gram

The molar mass of oxygen gas = 32 g/mol

We need to find the number of moles of oxygen. We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{34.4\ g}{32\ g/mol}\\\\=1.075\ mol

So, there are 1.075  moles in 34.4 g of Oxygen gas.

6 0
2 years ago
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