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noname [10]
3 years ago
11

A 49.7 mL sample of gas in the cylinder is warm from 20°C to 92°C. What is the volume of the final temperature

Chemistry
1 answer:
Xelga [282]3 years ago
6 0

Answer:

61.9mL

Explanation:

Given parameters:

Initial volume  = 49.7mL

Initial temperature  = 20°C  = 293K

Final temperature  = 92°C  = 365K

Unknown:

Final volume  = ?

Solution:

To solve this problem, we simply apply the Charles's law which states that "the volume of a fixed mass of a gas varies directly as it absolute temperature if the pressure is  constant";

        \frac{V1}{T1}   = \frac{V2}{T2}  

V and T are volume and temperature values

1 and 2 are initial and final states

 Now insert the parameters and solve;

      \frac{49.7}{293}   = \frac{V2}{365}  

        V2  = 61.9mL

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How many moles of ammonia can be formed from 4.0 mol H2
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Answer: 1. 2.7 moles  of ammonia are formed

2. 12.0 moles  of hydrogen are required

3. 2.0 moles of nitrogen are required

Explanation:

The balanced chemical equation is:

N_2+3H_2\rightarrow 2NH_3

According to stoichiometry:

3 moles of hydrogen form = 2 moles of ammonia

Thus 4.0 moles of hydrogen form =\frac{2}{3}\times 4.0=2.7moles of ammonia

According to stoichiometry:

2 moles of ammonia are formed by = 3 moles of hydrogen

Thus 8.0 moles of ammonia are formed by  =\frac{3}{2}\times 8.0=12.0moles of hydrogen

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3 years ago
Explain how intermolecular attractions between molecules influence the bulk of properties of a material
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Answer:

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5 0
2 years ago
Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)     \Delta H_1=-241.8kJ

(2) X(s)+2Cl_2(g)\rightarrow XCl_4(s)    \Delta H_2=+461.9kJ

(3) \frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)    \Delta H_3=-92.3kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) H_2O(g)\rightarrow H_2O(l)    \Delta H_5=-44.0kJ

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) 2H_2O(g)\rightarrow 2H_2(g)+O_2(g)     \Delta H_1=2\times 241.8kJ=483.6kJ

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(3) 2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)    \Delta H_3=4\times -92.3kJ=-369.2kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) 2H_2O(l)\rightarrow 2H_2O(g)    \Delta H_5=2\times 44.0kJ=88.0kJ

The expression for enthalpy of main reaction will be:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5

\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)

\Delta H=-1048.6kJ

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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Hope i helped... If you need anything else ask me! :)

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