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aivan3 [116]
2 years ago
3

The recipe for pumpkin pie calls for 3 tablespoons of flour and 1.5 cups of sugar for each pie. Which of the following conversio

n factors would be used to find out how many pies could be made from 7.5 cups of sugar.
Chemistry
1 answer:
Gwar [14]2 years ago
7 0

Answer:

5 pumpkin pies could be made with 7.5 cups of sugar.

Explanation:

The conversion factor to solve the problem is:

7.5 cups of sugar\times\frac{1 pie}{1.5 cups of sugar} = 5 pies

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The first step in the Ostwald process for producing nitric acid is 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g). If the reaction of
wel

Answer:

a. 77%

Explanation:

To solve the exercise, you need to know the limiting reagent. The limiting reagent is one that is consumed first in the chemical reaction. The other reagent is known as an excess reagent, and it is the one left over in a chemical reaction.

To know if ammonia or oxygen is the limiting reagent, we should relate the weights of both compounds in the reaction. We know that 4NH3 (68 g) react with 5O2 (160 g), so we should find out how much NH3 is required to react with 150 g of O2:

160 g O2 _______ 68 g NH3

150 g O2 _______ x = 150 g * 68 g / 160 g = 63.75 g NH3

As we have 150 g of NH3 but only 63.75 g are required, we know that ammonia is the excess reagent, so oxygen is the reagent that limits the reaction and that we should use to calculate the percentage yield of the reaction.

In the reaction we observe that 5O2 (160 g) produces 4NO (120g), so:

160 g O2 ______ 120 g NO

150 g O2 ______ x = 150 g * 120 g / 160 g = 112.5 g NO

If the percent yield of the reaction were 100%, it would produce 112.5 g of NO, however only 87 g of NO were obtained, so we should find out what the percentage of reaction yield was:

112.5 g NO ______ 100%

87 g NO _______ x = 87g * 100% / 112.5 g = 77%

Thus we observe that the reaction had a yield of 77%.

7 0
2 years ago
Read 2 more answers
C5H12, pentane, is a liquid at room temperature.
Art [367]

Answer:

C)We cannot be sure unless we find out its boiling point.

Explanation:

It is necessary to clearly explain here that simply observing two compounds of the same homologous series irrespective of how close they may be in the series will not give us the faintest idea regarding which one will be a liquid, solid or gas at room temperature.

However, to determine whether an unknown substance will be a liquid at room temperature, then its important to measure its boiling point. If the boiling point is above room temperature, and the melting point is below room temperature, the compound is a liquid. If the boiling point of the unknown substance is below room temperature, it is a gas.

It is now safe to conclude that cannot decide on the state of matter in which a compound exists unless we know something about its boiling point, not merely looking closely at the properties of its neighbouring compounds in the same homologous series

8 0
2 years ago
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°C _____.
MatroZZZ [7]

Answer:

c

Explanation:

1 calorie = 4.184J/g×°C

This also happens to be the specific heat capacity of water, which is the amount of energy it takes to raise the temperature of 1mL of water by 1°C

6 0
2 years ago
A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h
Ksenya-84 [330]

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution : 0.0328 M

Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}

Moles of HCl in 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution : 0.0245 M

Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}

Moles of NaOH in 0.100 L solution = 0.00245 moles

3) Concentration of hydrochloric acid in the resulting solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.

Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L

Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles =  0.00575 moles

Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}

Molarity of HCl left un-neutralized :\frac{0.00575 moles}{0.350L}=0.0164 M

0.0164 molar concentration of hydrochloric acid in the resulting solution.

3 0
2 years ago
Compare Solution X to Solution A, Solution B, and Solution C. Beaker with a zoom view showing eight shperes, four groups of part
Levart [38]

Answer:

SOLUTION A

Explanation:

7 0
2 years ago
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