Chemistry work here. Please help as soon as possible. I have allot of questions that needs to be answered. Can someone do it for me?

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Answer 1

Answer:

Answer:

I dont know about the yield but I think mno2 is the limiting reactant

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I will give Brainliest and 30 points to whoever answers first!

From which type of clouds do hail pellets form?

Answers

Answer:

Hail pallets forms inside of cumulonimbus clouds

Explanation:

Answer: cumulonimbus hope it helps pls stay safe

Which element has a complete valence electron shell?selenium (Se)
oxygen (0)
fluorine (F)
argon (Ar)
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Final answer:

Argon (Ar) has a complete valence electron shell.

Explanation:

The element that has a complete valence electron shell is argon (Ar).

Learn more about complete valence electron shell here:

brainly.com/question/2725863

The solubility of glucose at 30°C is125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

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Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL * (0.996g)/(1mL) =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O * (550gGlucose)/(398gH_2O) = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated. If you want to increase the amount of glucose in the solution without adding more glucose, you can increase the temperature.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated.

Since the solubility of glucose at 30°C is 125 g/100 g water, adding 550 g of glucose to 400 mL of water exceeds the maximum amount of glucose that can dissolve in the given amount of water.

To increase the amount of glucose in the solution without adding more glucose, you can increase the temperature. Higher temperatures generally increase the solubility of solutes in water. By increasing the temperature, you can dissolve more glucose in the solution.

Learn more about solubility here:

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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 50.0L tank with 14. mol of sulfur dioxide gas and 2.6 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.6 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

Answers

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

$2SO_2 + O_2 \to 2SO_3$

The I.C.E table can be represented as:

2SO₂ O₂ 2SO₃

Initial: 14 2.6 0

Change: -2x -x +2x

Equilibrium: 14 - 2x 2.6 - x 2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction $2SO_2 + O_2 \to 2SO_3$ = 0.8325;

If we want to find:

$SO_2 + (1)/(2)O_2 \to SO_3$

Then:

$K_c = (0.8325)^(1/2)$

$\mathbf{K_c = 0.912}$

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

MassWhat is the percent composition by mass of
nitrogen in the compound N,H, (gram-formula
32 g/mol)?
(1) 13%
(3) 88%
(2) 44%
(4) 93%

Answers

Answer: 88%

Explanation:

Solve the problems. Express your answers to the correct number of significant figures. 0.09 + 1.324 =
41.0 + 78.3 =

Answers

Ans: 1) 1.41

2) 119.3

While adding two numbers, the rule of significant figures requires that the number of decimal places in the final answer is equal to that of the term with the least decimal places

In the given examples:

1) 0.09 - has 2 decimal places

1.324- has 3 decimal places

Sum = 0.09 + 1.324 = 1.414

Round off to 2 decimal places = 1.41

2) 41.0 - has 1 decimal place

78.3- has 1 decimal place

Sum = 41.0 + 78.3 = 119.3

Final answer will also have 1 decimal place = 119.3

1. 1.141

2. 119.3

HOPE THIS HELPS.