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Softa [21]
3 years ago
10

How many moles are there in 40g of calcium

Chemistry
1 answer:
drek231 [11]3 years ago
8 0
40 grams ÷ 40.08 grams/moles = 1 mole
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ki77a [65]

Answer:

We'll have 82 moles ZnO and 41 moles S

Explanation:

Step 1: data given

Number of moles Zinc (Zn) = 82 moles

Number of moles sulfur oxide (SO2) = 42 moles

Step 2: The balanced equation

2Zn + SO2 → 2ZnO + S

Step 3: Calculate the limiting reactant

For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur

Zinc is the limiting reactant. It will completely be consume (82 moles). Sulfur oxide is in excess. There will react 82/2 = 41 moles

There will remain 42-41 = 1 mol SO2

Step 4: Calculate moles of products

For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur

For 82 moles Zinc we'll have 82 moles of Zinc Oxide (ZnO)

For 82 moles Zinc we'll have 82/2 = 41 moles of sulfur

We'll have 82 moles ZnO and 41 moles S

5 0
3 years ago
Use Target Reading Skills After you read the section, reread the paragraphs that contain definitions of Key Terms. Use all of th
Ksju [112]
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3 0
2 years ago
A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1
icang [17]

Answer:

Approximately 0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}.

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C.

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took 2911\; \rm J of energy to raise the temperature of this sample by \Delta T = 92.6\; \rm ^\circ\! C. Therefore, raising the temperature of this sample by 1\; \rm ^\circ\! C (unit temperature) would take only \displaystyle \frac{1}{92.6} as much energy. That corresponds to approximately 31.436\; \rm J of energy.

On the other hand, the energy required to raise the temperature of this material by 1\; \rm ^\circ\! C is proportional to the mass of the sample (also assuming no phase change.)

It took approximately 31.436\; \rm J of energy to raise the temperature of 57.07\; \rm g of this material by 1\; \rm ^\circ C. Therefore, it would take only \displaystyle \frac{1}{57.07} as much energy to raise the temperature of 1\; \rm g (unit mass) of this material by 1\; \rm ^\circ \! C\!. That corresponds to approximately 0.551\; \rm J of energy.

In other words, it takes approximately 0.551\; \rm J to raise 1\; \rm g (unit mass) of this material by 1\; \rm ^\circ \! C. Therefore, by definition, the specific heat of this material would be approximately 0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}.

8 0
2 years ago
Which volume is the greatest?
Zolol [24]
<span>centiliter is the correct answer</span>
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3 years ago
Weak tea is an example of a...
podryga [215]
It is a dilute solution

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