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eimsori [14]
3 years ago
6

Which sentence best explains the relationship between pressure and the

Chemistry
1 answer:
ella [17]3 years ago
3 0

Answer: A. The greater the pressure, the more of gas that will dissolve.

Explanation: Increasing pressure increases the solubility of gases. It has little effect on the solubility of liquids and solids.

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By conduction from your hand to the glass. By convection from your hand to the glass. By convection from the ice to your hand . By radiation from the ice to your hand.
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A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC
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Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 25^oC=273+25=298K

V = volume = 4.0 L

n_1 = moles of SO_2 = 0.20 mol

n_2 = moles of O_2 = 0.20 mol

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

P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}

P=2.4atm

Thus, the pressure in the flask after reaction complete is, 2.4 atm

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3 years ago
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How many grams of kclo3 are needed to make 30.0 grams of kcl
lubasha [3.4K]

The mass of  potassium chlorate, KClO₃ needed to produce 30 grams of potassium chloride, KCl is 49.33 grams

<h3>Balanced equation </h3>

2KClO₃ —> 2KCl + 3O₂

Molar mass of KClO₃ = 39 + 35.5 + (16×3) = 122.5 g/mol

Mass of KClO₃ from the balanced equation = 2 × 122.5 = 245 g

Molar mass of Kcl= 39 + 35.5 = 74.5 g/mol

Mass of KCl from the balanced equation = 2 × 74.5 = 149 g

SUMMARY

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

<h3>How to determine the mass of KClO₃ needed </h3>

From the balanced equation above,

149 g of KCl were obtained from 245 g of KClO₃

Therefore,

30 g of KCl will be obtained from = (30 × 245) / 149 = 49.33 g of KClO₃

Thus, 49.33 g of KClO₃ are needed for the reaction

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