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Artemon [7]
3 years ago
12

A gas at constant temperature has a pressure of 404.6 kPa with a volume of 12 ml. If the volume changes to 43ml, what is the new

pressure
Chemistry
1 answer:
blagie [28]3 years ago
4 0

Answer:

The answer is

<h2>112.912 kPa</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

404.6 kPa = 404600 Pa

From the question we have

P_2 =  \frac{404600 \times 12}{43}  =  \frac{4855200}{43}  \\  = 112911.6279... \\  = 112912

We have the final answer as

<h3>112.912 kPa</h3>

Hope this helps you

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An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of
kherson [118]

Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

MAss glucose = moles glucose* molecular weight of glucose

MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

We need 8.11 grams of glucose for this solution

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3 years ago
How many orbitals are in the n=3 level
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It shows the atomic number
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If something is made of atoms it must be what?
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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
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<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.

4 0
3 years ago
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