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lbvjy [14]
3 years ago
12

How many moles are there in 814.504 g of H2O?

Chemistry
1 answer:
earnstyle [38]3 years ago
6 0
First you need to calculate the molar mass of H2O. To do that, look at the periodic table and add up the AMUs (atomic mass units) 

H = 1.01
O = 16.0 

1.01 • 2 + 16.0 = 18.02 

Next, use stoichiometry to convert grams into moles. To do that, divide 814.504g by the number of grams in one mole of H2O.

814.504 ÷ 18.02 = 45.2 moles 

There are 45.2 moles in 814.504 grams of H2O.
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If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,
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Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

\frac{P1V1}{RT1} = \frac{P2V2}{RT2}

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

\frac{1*20}{R*10} = \frac{10*V2}{R*100}

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

\frac{1*20*100}{10*10} = V2

∴ V2 = \frac{2000}{100}

∴ V2 = 20mL

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What mass of aluminum chloride could be made from 8.1 g of aluminum and 4.2 L of chlorine at STP?
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When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
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<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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