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2 years ago

What would you observe in a solution of 30 g of KC1O3 in 100 g of water at 10 c?

Chemistry

1 answer:

Mrac [35]2 years ago

7 0

Answer:

Option A is correct. About 5 g of the KClO3 is dissolved

Explanation:

KClO3 is not very good soluble in water.

So, Option C is impossible, because KClO3 is poorly soluble in water.

The low solubility of KClO3 in water causes KClO3 to isolate itself from the reaction mixture by precipitating out of solution.

So, option D will either happen.There will be a part of KClO3 dissolve.

At 10 °C, KClO3 has a solubility of 4.46 g/100 gram (10 °C).

Option A is correct. About 5 g of the KClO3 is dissolved

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spin [16.1K]

Answer:

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2 years ago

Read 2 more answers

Solve the ideal gas law equation for pressure.

posledela

Answer:

$p=\frac{nRT}{V}$

Explanation:

The ideal gas law equation is an equation that relates some of the quantities that describe a gas: pressure, volume and temperature.

The equation is:

$pV=nRT$

where

p is the pressure of the gas

V is the volume of the gas

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas (must be expressed in Kelvin)

Here we want to solve the equation isolating p, the pressure of the gas.

We can do that simply by dividing both terms by the volume, V. We find:

$p=\frac{nRT}{V}$

So, we see that:

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2 years ago

What actions affect blood pH

galben [10]

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2 years ago

What is the molarity (M) of the following solutions?

Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

= 27 + 3(16 + 1)

= 27 + 3(17) = 27 + 51

= 78 g/mole

$Al(OH)_3$ = 78 g/mole

Given mass= 19.2 g/mole

$Mole = \frac{Given\ mass}{Molar\ mass}$

$Mole = \frac{19.2}{78}$

Moles = 0.246

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

Volume = 280 mL = 0.280 L

$Molarity = \frac{0.246}{0.280)}$

Molarity = 0.879 M

Molarity = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

$Mole = \frac{235.9}{119}$

Moles = 1.98

Volume = 2.6 L

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

$Molarity = \frac{1.98}{2.6)}$

Molarity = 0.762 M

Molarity = 0.76 M

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2 years ago

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olga_2 [115]

Answer:

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Explanation:

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3 years ago