What is the mass of phosphorus that contains twice the number of atoms found in 14
g of iron?
[Relative atomic mass : P = 31; Fe = 56]

Answers

Answer 1

Answer:

15.5 gm

Explanation:

What is the mass of phosphorus that contains twice the number of atoms found in 14 g of iron?

[Relative atomic mass : P = 31; Fe = 56]

14 gm Fe = 14gm/ 56 gm/mole = 14 mole gm/56gm = 14/56 mole

0.25 moles

2 X 0.25 = 0.5 moles

1 mole P = 31 gm

so

0.5 moles P =31/2 =15.5 gm


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Explanation:

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Explanation:

Main compounds of milk are lactose and casein. And it is also called a colloidal mixture (i.e. in which one substance of microscopically dispersed insoluble or soluble particles is suspended throughout another substance). Therefore milk is regarded as a mixture not as a pure substance. So this is the required answer.

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Answer:

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Answer:

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Explanation:


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Answers

Answer:

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Explanation:

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Hope this helps :)

đơn chất nào sau đây tác dụng với dung dịch axit sunfuric loãng sinh ra khí ?
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Answer:

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Answer:

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Explanation:

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Answer:

It's because the solubility of gases in water is inversely proportional to the temperature, so it means when the temperature is increased, the solubility decreases, and the gases tend to get out from the soda

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Answers

The effects of long-term exposure to mustard gas causes mortality and chronic respiratory disease.

Exposure to mustard gas in the long term can result in mortality from influenza, pneumonia as well as chronic breathing disease. Exposure for long term leads to respiratory complications and septic shock. Extensive breathing in the environment of mustard gas can cause chronic respiratory disease, respiratory infections or death.

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Answer:

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I think it is helps you

Please help I will give brainliest ;3

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Answer:

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ANSWER

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0
4.00 am
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Answers

A 3.00-L container with O₂ at 4.00 atm is connected to a 2.00-L container with Ne at 2.00 atm, resulting in a final pressure of 3.20 atm in the container.

We have 2 containers connected by a valve.

One container has a volume of 3.00 L and contains O₂ at 4.00 atm. The other container has a volume of 2.00 L and contains Ne at 2.00 atm.

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O₂

[tex]P_1 \times V_1 = P_2 \times V_2\\\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{4.00atm \times 3.00L}{5.00L} = 2.40atm[/tex]

Ne

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The final pressure in the container will be the sum of the final pressures of the gases.

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Answer:

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superscript Coefficient

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Answers

Answer:

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