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mylen [45]
4 years ago
10

A 642 mL sample of oxygen gas at 23.5°C and 795 mm Hg, is heated to 31.7°C and the volume of the gas expands to 957 mL. What is

the new pressure in atm?
A) 0.683 atm
B) 0.932 atm
C) 3.05 atm
D) 5.19 atm
Chemistry
1 answer:
Alexus [3.1K]4 years ago
3 0
The correct answer is approximately 0.7 atm, which is close to A) 0.683 atm.
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Evgen [1.6K]

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c.the speed of the bail decrease over time

6 0
3 years ago
Read 2 more answers
Positive oxidation number suggest that the element wants to ____ (lose/gain) electrons?
BARSIC [14]

Answer:

lose

Explanation:

its lose because if its positive, assume the electron is a person who has already eaten, theyre full and dont want more, therefore they would want to lose the weight, so its lose.

8 0
3 years ago
PLEASE HELP CHEM BABES I HAVE BEEN CRYING FOR A WHILE NOW
slavikrds [6]

Answer: See below

Explanation:

1. To calculate the mass, you know you can convert by using molar mass. Since mass is in grams, we can use molar mass to convert moles to grams. This calls for the Ideal Gas Law.

Ideal Gas Law: PV=nRT

We manipulate the equation so that we are solving for moles, then convert moles to grams.

n=PV/RT

P= 100 kPa

V= 0.831 L

R= 8.31 kPa*L/mol*K

T= 27°C+273= 300 K

Now that we have our values listed, we can plug in to find moles.

n=\frac{(100kPa)(0.831L)}{(8.31kPa*l/molK)(300K)}

n=0.033mol

We use the molar mass of NO₂ to find grams.

0.033mol*\frac{46.005g}{1mol }=1.52 g

The mass is 1.52 g.

2. To calculate the temperature, we need to use the Ideal Gas Law.

Ideal Gas Law: PV=nRT

We can manipulate the equation so that we are solving for temperature.

T=PV/nR

P= 700.0 kPa

V= 33.2 L

R= 8.31 kPa*L/mol*K

n= 70 mol

Now that we have our values, we can plug in and solve for temperature.

T=\frac{(700kPa)(33.2L)}{(70mol)(8.31 kPa*L/molK)}

T=40K

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3 0
3 years ago
You are given a piece of paper and a match. The paper has a mass of 2.5 g. You then light the match and light the piece of paper
kow [346]

Answer:

No

Explanation:

<em>No. </em>T<em>he demonstration does not violate the conservation of mass.</em>

<u>The law of conservation of mass states that mass can neither be created nor destroyed in a reaction. However, mass can be converted from one form to another during the reaction.</u>

In this case, even though the remaining bits of paper weigh 0.5 g while the original paper weighed 2.5 g, the ashes and smoke/gas from the burning will all add up to the lost weight of the paper.

<em>The burned part has been converted into other forms. If the smoke/gas and the ashes are properly captured, they will mark up with the weight of the remaining paper to give the weight of the original paper. </em>

5 0
3 years ago
What is the energy change that will occur when 25.5 grams of oxygen gas (O2) react with excess methane (CH4) according to the re
AysviL [449]
Amount of CH4 is excess, so no need to worry about it 
<span>but the limiting factor is the Oxygen </span>

<span>according to stranded equation, </span>

<span>CH4 + 2 O2 --> CO2 + 2 H2O ΔH = -889 kJ/mol </span>


<span>just by taking proportions </span>

<span>(-889 kJ/mol) / 2 x 0.8 mol = - 355.6 kJ </span>

<span>so i think the answer is (a)</span>
7 0
3 years ago
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