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3 years ago

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

Bess [88]

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

According to stoichiometry:

3 moles of hydrogen react with = 1 mole of nitrogen

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3 years ago

Explain how intermolecular attractions between molecules influence the bulk of properties of a material

irina1246 [14]

Answer:

Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid.

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

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

(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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3 years ago

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ludmilkaskok [199]

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3 years ago

What Is a decrease in velocity called

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The decrease in velocity is called deceleration or negative acceleration.

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

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