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Ugo [173]
2 years ago
15

What quantity of energy is required to heat a piece of iron wighing 1.31 g from 25.0 degrees celsius to 46.0 degrees celsius

Chemistry
1 answer:
Ratling [72]2 years ago
4 0
The mass energy of telegranium
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Prove that PV = nRT.​
qaws [65]

Find your answer in the explanation below.

Explanation:

PV = nRT is called the ideal gas equation and its a combination of 3 laws; Charles' law, Boyle's law and Avogadro's law.

According to Boyle's law, at constant temperature, the volume of a gas is inversely proportional to the pressure. i.e V = 1/P

From, Charles' law, we have that volume is directly proportional to the absolute temperature of the gas at constant pressure. i.e V = T

Avogadro's law finally states that equal volume of all gases at the same temperature and pressure contain the same number of molecules. i.e V = n

Combining the 3 Laws together i.e equating volume in all 3 laws, we have

V = nT/P,

V = constant nT/P

(constant = general gas constant = R)

V = RnT/P

by bringing P to the LHS, we have,

PV = nRT.

Q.E.D

6 0
2 years ago
An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the
Misha Larkins [42]

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

3 0
3 years ago
Two new planets with equal masses are discovered. Planet A is twice as far from the sun as Planet B. Which of the following is t
Sphinxa [80]
I think the answer is <span>B. The sun's gravitational pull on Planet A is equal to its gravitational pull on Planet B because distance does not affect gravity.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
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7 0
3 years ago
Read 2 more answers
Explain how you would convert the moles of a compound to the made (grams) of the same compound<br>​
Oduvanchick [21]

Answer:

see notes below

Explanation:

The mole is the mass of substance containing 1 Avogadro's Number of particles. That is, 1 mole substance = 1 formula weight. For elements, 1 mole weight is equal to the atomic weight expressed as grams. For molecules, 1 mole weight is equal to the molecular weight expressed as grams.

1 mole = 1 formula weight

<u>Moles to Grams and Grams to Moles</u>

Grams => Moles

Given grams, moles = mass given / formula weight

*Ask the question => How many formula weights are there in the given mass? => Results is always moles.

Moles => Grams

Given moles,  grams = moles given X formula weight

*Summary

Grams to Moles => divide by formula weight

Moles to Grams => multiply by formula weight

3 0
3 years ago
How many moles of N2O5 are needed<br> to produce 7.90 g of NO2?
Roman55 [17]

Answer:

0.085 moles of  N₂O₅ are needed

Explanation:

Given data:

Mass of NO₂ produces = 7.90 g

Moles of N₂O₅ needed = ?

Solution:

2N₂O₅       →     4NO₂  + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

Number of moles = 7.90 g/ 46 g/mol

Number of moles = 0.17 mol

now we will compare the moles of NO₂   with N₂O₅.

                NO₂          :       N₂O₅

                  4            :          2

                0.17          :         2/4×0.17 = 0.085 mol

Thus, 0.085 moles of  N₂O₅ are needed.

4 0
2 years ago
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