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3 years ago

What is the volume of 43.7 g of helium at stp?

2 answers:

0 0

Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
V(He) = 10.925 mol · 22.4 L/mol.
V(He) = 244.72 L.
Vm - molar volume at STP.
n - amount of substance.

egoroff_w [7]3 years ago

0 0

Answer : The volume of helium gas is, 244.72 liters

Explanation : Given,

Mass of helium gas = 43.7 g

Molar mass of helium gas = 4 g/mole

As we know that at STP, 1 mole of gas contains 22.4 liter volume of gas.

First we have to calculate the moles of helium gas.

$\text{Moles of He}=\frac{\text{Mass of He}}{\text{Molar mass of He}}=\frac{43.7g}{4g/mole}=10.925mole$

Now we have to calculate the volume of helium gas.

As, 1 mole of helium gas contains 22.4 liter volume of gas

So, 10.925 mole of helium gas contains $10.925\times 22.4=244.72$ liter volume of gas

Therefore, the volume of helium gas is, 244.72 liters

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3 years ago

Calculate the amount of energy (in kilojoules) needed to heat 346 g of liquid water from –10 °C to 182°C. Assume that the specif

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Answer : The amount of heat required is, $1.16\times 10^6J$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(182^oC)$

The expression used will be:

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat required for the reaction

m = mass of ice = 346 g

$c_{p,s}$ = specific heat of solid water or ice = $2.09J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.184J/g^oC$

$c_{p,g}$ = specific heat of gaseous water = $1.99J/g^oC$

$\Delta H_{fusion}$ = enthalpy change for fusion = $333J/g$

$\Delta H_{vap}$ = enthalpy change for vaporization = $2260J/g$

Now put all the given values in the above expression, we get:

$\Delta H=[346g\times 2.09J/g^oC\times (0-(-80))^oC]+346g\times 333J/g+[346g\times 4.184J/g^oC\times (100-0)^oC]+346g\times 2260J/g+[346g\times 1.99J/g^oC\times (180-100)^oC]$

$\Delta H=1.16\times 10^6J$

Therefore, the amount of heat required is, $1.16\times 10^6J$

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3 years ago

!!!!HURRY!!!! What reaction is endothermic?

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I think it's answer is 2nd option.

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4 years ago