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Rama09 [41]
4 years ago
7

Given the following balanced chemical equation:

Chemistry
1 answer:
Fudgin [204]4 years ago
7 0

Answer:

Mass of Ca(OH)₂ required  = 0.09 g  

Explanation:

Given data:

Volume of HNO₃ = 25 mL (25/1000 = 0.025 L)

Molarity of HNO₃ = 0.100 M

Mass of Ca(OH)₂ required  =  ?

Solution:

Chemical equation;

Ca(OH)₂ + 2HNO₃   →      Ca(NO)₃ + 2H₂O

Number of moles of HNO₃:

Molarity = number of moles / volume in L

0.100 M = number of moles / 0.025 L

Number of moles = 0.100 M ×0.025 L

Number of moles = 0.0025 mol

Now we will compare the moles of Ca(OH)₂  with HNO₃ from balance chemical equation.

                   HNO₃           :           Ca(OH)₂    

                       2               :              1

                  0.0025         :            1/2×0.0025 = 0.00125

Mass of Ca(OH)₂:    

Mass = number of moles × molar mass

Mass =   0.00125 mol × 74.1 g/mol

Mass = 0.09 g    

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Hydrogen + oxygen -> water

2H2 + O2 -> 2H2O (balanced)

2 moles of hydrogen gas will react with 1 mole of oxygen to produce 2 moles of water.

following that ratio, 0.734 mole of oxygen must need 0.734 × 2 moles of hydrogen to fully react.

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6 0
4 years ago
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If ammonia is manufactured at 356 k, is the reaction spontaneous, given that the enthalpy and entropy change for the reaction ar
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The reaction will be spontaneous if Gibb free energy is negative, according the following relation:  
G = H - T*S  
where G is Gibbs free energy, T is change of enthalpy , T is temperature and S is change of entropy  
In the case of ammonia:  
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4 years ago
Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0
klemol [59]

Answer:

Partial pressure of methane: 1.18 atm

Partial pressure of ethane: 1.45 atm

Partial pressure of propane: 2.35 atm

Explanation:

Let the total moles of gases in a container be n.

Total pressure of the gases in a container =P = 5.0 atm

Temperature of the gases in a container =T = 23°C = 296.15 K

Volume of the container = V = 10.0 L

PV=nRT (Ideal gas equation)

n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol

Moles of methane gas =n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol

Moles of ethane gas =n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol

Moles of propane gas =n_3=?

n=n_1+n_2+n_3

n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol

Partial pressure of all the gases can be calculated by using Raoult's law:

p_i=P\times \chi_i

p_i = partial pressure of 'i' component.

\chi_1 = mole fraction of 'i' component in mixture

P = total pressure of the mixture

Partial pressure of methane:

p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}

p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm

Partial pressure of ethane:

p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}

p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm

Partial pressure of propane:

p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}

p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm

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Answer:

The following three isomeric structure are given below.

Explanation:

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Ester #1: methyl 1-methylcyclobutanecarboxylate

Ester #2: (E)-methyl 3-methyl-3-pentenoate

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Formation of hydrates is considered to be a chemical change.

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