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4 years ago

How many molecules of H20 are in 2.0 moles

Chemistry

1 answer:

inessss [21]4 years ago

4 0

Hi There! :)

<span>How many molecules of H20 are in 2.0 moles
</span>
12.044*10^23 or 6.022⋅1023 <span>atoms</span>

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tia_tia [17]

Answer: 10.2 grams

Explanation:

The balanced chemical reaction is :

$CaH_2(s)+2H_2O(l)\rightarrow Ca(OH)_2(aq)+2H_2(g)$

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 740 torr = 0.97 atm (760torr=1atm)

V= Volume of the gas = 12.0 L

T= Temperature of the gas = 19°C = 292 K $0^0C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas

$n=\frac{PV}{RT}=\frac{0.97\times 12.0}{0.0821\times 292}$

$n=0.48$

According to stoichiometry:

2 moles of hydrogen are generated by = 1 mole of $CaH_2$

Thus 0.48 moles of hydrogen are generated by = $\frac{1}{2}\times 0.48=0.24$ moles of $CaH_2$

Mass of $CaH_2=moles\times {\text {Molar mass}}=0.24mol\times 42g/mol=10.2g$

Thus 10.2 grams of $CaH_2$ are needed to generate 12.0 L of hydrogen gas if the pressure of hydrogen is 740. torr at 19°C

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The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc

jek_recluse [69]

Answer: $K_{sp}=1.25\times 10^{-14}$

$pK_{sp}=13.90$

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$Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}$

When the solubility of $Mg_3(PO_4)_2$ is S moles/liter, then the solubility of $Mg^{2+}$ will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.