If a saturated solution of potassium chloride is cooled from 80℃ to 50℃,how much precipate is formed

Answers

Answer 1

Answer: Roughly 39 grams, give or take 1 gram

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What volume will 20.0g of Argon occupy at STP?

Answers

Molar mass Argon = 39.948 g/mol

1 mol ------ 39.948 g
mol ----- 20.0 g

mol = 20.0 * 1 / 39.948

= 0.5006 moles

1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L

L = 0.5006 * 22.4

= 11.21 L

hope this helps!

In the following combustion reaction of acetylene (C2H2), how many liters of CO2 will be produced if 60 liters of O2 is used, given that both gases are at STP?2C2H2+5O2 2H2O+4CO2

The volume of one mole of gas at STP is 22.4 liters.

Answers

The stoichiometry of the reaction is 5 mols of O2 produces 4 mols of CO2.

1 mol at STP is equivalent to 22.4 liters.

So, 5*22.4 liters of O2 produces 4*22.44 liters of CO2

Then 60 liters of O2 produces 60*4/5 = 48 liters of CO2

Final answer:

In the combustion reaction of acetylene (C2H2), 60 liters of CO2 will be produced if 60 liters of O2 is used.

Explanation:

To determine the volume of CO2 produced in the combustion reaction of acetylene (C2H2), we need to use stoichiometry. From the balanced equation, we can see that 2 moles of C2H2 produce 4 moles of CO2. The ratio is 2:4 or 1:2. Given that 60 liters of O2 is used, we can assume the same volume of CO2 will be produced since both gases are at STP.

Therefore, the volume of CO2 produced would be 60 liters as well.

Keywords: combustion, volume, acetylene, O2, CO2, STP

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A gas changes directly to a solid during ....fusion
saponification
depositon
decomposition

Answers

The answer is Deposition.

What did Thomson’s model of the atom include that Dalton’s model did not have?

Answers

Thomson's model included Protons and Electrons. His model is referred to as 'Plum Pudding' because of it.

Answer: subatomic particles: negative charges (electrons) distributed in a mass of positive charge.

Explanation:

1) John Dalton's model depicted the matter as the combination of tiny, indivisible particles, called atoms.

According to this model, atoms can not be created, destroyed, or divided into smaller particles.

2) When it was discovered that all forms of matter contained negative particles, by multiple experiments with cathode ray tubes, those particles where named electrons.

3) J.J. Thompson could determine that the mass of those negative charges was much smaller that the mass of the smallest atom (hydrogen). Concluding that existed smaller particles than the atom. Hence, Dalton's model was wrong: atoms was divisible into smaller subatomic particles.

4) Then J.J Thompson proposed the plum pudding model, in which the electrons (plums) are embeded into a uniform positive mass (pudding).

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.391 g of carbon dioxide is produced from the reaction of 0.16 g of methane and 0.84 g of oxygen gas, calculate the percent yield of carbon dioxide.

Answers

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=(0.16g)/(16g/mole)=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(0.84g)/(32g/mole)=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $(0.026)/(2)=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $(0.026)/(2)=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2* \text{ Molar mass of }CO_2$