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SVETLANKA909090 [29]
3 years ago
14

Wait what's the answer? What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?

Chemistry
1 answer:
PolarNik [594]3 years ago
7 0
B. Definitely B. Hope this helps :3

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In the food chain what might happen if all the frogs suddenly died off?
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The food chain will crash.
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In which material are the particles arranged in a regular geometric pattern?
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 I think only solids arrange in a regular geometric pattern, so i2
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Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0
3 years ago
Which of the following is defined as the mass of an atom based on the mass of an atom of carbon-12?
iogann1982 [59]
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3 0
3 years ago
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Baking soda (NaHCO3) and vinegar (HC2H3O2) react to form sodium acetate, water, and carbon dioxide. If 42.00 g of baking soda re
Setler [38]

Answer:

0.5 mole of CO₂.

Explanation:

We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:

Mass of NaHCO₃ = 42 g

Molar mass of NaHCO₃ = 23 + 1 + 12 + (16×3)

= 23 + 1 + 12 + 48

= 84 g/mol

Mole of NaHCO₃ =?

Mole = mass / molar mass

Mole of NaHCO₃ = 42/84

Mole of NaHCO₃ = 0.5 mole

Next, balanced equation for the reaction. This is given below:

NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂

Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.

Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.

Thus, 0.5 mole of CO₂ was obtained from the reaction.

7 0
3 years ago
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