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Gre4nikov [31]
3 years ago
5

A gas sample has a pressure of 2.65 atm when the temperature is -18 ∘C. What is the final pressure, in atmospheres, when the tem

perature is 39 ∘C, with no change in the volume or amount of gas?

Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

P₂ = 3.24atm

Explanation:

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Why is a neutral iron atom a different element than a neutral carbon atom?
katovenus [111]
<span>Atomic number. In all atoms the distinct properties of that atom are a result of the number of protons that atom is composed of. A carbon atom has six protons, while an iron atom has 26.</span>
3 0
3 years ago
What is the ratio of 6 triangles to 4 circles?
harina [27]

Answer:six tryangles to four circle

Explanation:yiu would say it just As it is

5 0
3 years ago
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Select the coefficients necessary to balance each equation. Choose a coefficient for every compound.
ivolga24 [154]

Answer:

See below

Explanation:

2C2H6 + 7O2 = 4CO2 + 6H2O

2C2H2 + 5O2 = 4CO2 + 2H2O

A lot of this is trial and error.  I start from left to right.  For the first reaction, you had

C2H6 + O2  =  CO2  +  H2O

You have 2 carbons on the left so you can try adding a 2 in front of the carbons on the right and keep going from there.  Now you have

C2H6  +  O2  =  2CO2  + H2O

You now have 2 C on each side.  C2 is the same as 2C.

The next step is to look at the hydrogens on the left.  You have 6 so you need 6 on the right.  You can add a 3 in front of the H on the right.  The H already has a subscript of a 2 so having the 3 in front will now make it 6H.  You now have this

C2H6  +  O2  =  2CO2  +  3H2O

So your C are equal (2 on each side) and your H is equal (6 on each side).  All that is left is the O.  So far you have 2 on the left but you have 7 on the right (4 from the 2CO2 and 3 more from 3H2O).  Since you have an even number on one side and an odd number on the other, there is no way to make this work.  So it's time to start over.

My next step was to put a 2 in front of C2H6 and keep going as we did above.  Start with this

2C2H6   + O2  = CO2  + H2O

You have 4 C on the left.  So you need to add a 4 in front of CO2 to get 4 C on the right.  Now you have this

2C2H6  +  O2  =  4CO2  +  H2O

Now on the left you have 12 H (2 in front times the 6 subscript on the H).  You need to have 12 on the right.  We can get that by putting a 6 in front of the H2O ( again the 6 in front times the 2 subscript of the H is 12 total).  You have this now

2C2H6  +  O2  = 4CO2  +  6H2O

You now have 4 C on both sides, 12 H on both sides.  All that is left now is the O.

On the right side, you have a total of 14 ( 8 from the 4CO2 and 6 from 6H2O).  So you'd need to add 7 in front of the O2 to make it 14.

You're done.  4 C on both sides, 12 H on both sides and 14 O on both sides.

The second equation I just did the same process.

6 0
3 years ago
The box in the above picture is falling from the top of a building to the ground. Two major forces are acting on the box as it f
MAVERICK [17]

Answer:

I think it's B

Explanation:

apologies if I get this wrong

8 0
3 years ago
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1. A sample of gold (Au) has a mass of 35.12 g.
Alecsey [184]

1a) Answer is: the number of moles of gold (Au) in the sample is 0.178 mol.

m(Au) = 35.12 g; mass of gold.

M(Au) = 196.97 g/mol; molar mass of gold.

n(Au) = m(Au) ÷ M(Au).

n(Au) = 35.12 g ÷ 196.97 g/mol.

n(Au) = 0.178 mol; amount of gold.

1b) Answer is: the number of atoms of gold (Au) is 1.073·10²³.

N(Au) = n(Au) · Na.

N(Au) = 0.178 mol · 6.022·10²³ 1/mol.

N(Au) = 1.073·10²³.

2a) Answer is: 0.0035 moles of sucrose.

m(C₁₂H₂₂O₁₁) = 1.202 g; mass of sucrose.

M(C₁₂H₂₂O₁₁) = 12 · Ar(C) + 22 · Ar(H) + 11 · Ar(O) · g/mol.

M(C₁₂H₂₂O₁₁) = 12 · 12.01 + 22 · 1.01 + 11 · 16 · g/mol.

M(C₁₂H₂₂O₁₁) = 342.3 g/mol; molar mass of sucrose.

n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).

n(C₁₂H₂₂O₁₁) = 1.202 g ÷ 342.3 g/mol.

n(C₁₂H₂₂O₁₁) = 0.0035 mol; amount of sucrose.

2b) n(C) = 12·n(C₁₂H₂₂O₁₁).

n(C) = 12 · 0.0035 mol.

n(C) = 0.042 mol; amount of carbon in sucrose.

n(H) = 22·n(C₁₂H₂₂O₁₁).

n(H) = 22 · 0.0035 mol.

n(H) = 0.077 mol; amount of hydrogen in sucrose.

n(O) = 11·n(C₁₂H₂₂O₁₁).

n(O) = 11 · 0.0035 mol.

n(O) = 0.0385 mol; amount of oxygen atoms in sucrose.

2c) N(C) = n(C) · Na.

N(C) = 0.042 mol · 6.022·10²³ 1/mol.

N(C) = 2.53·10²²; number of carbon atoms in sucrose.

N(H) = n(H) · Na.

N(H) = 0.077 mol · 6.022·10²³ 1/mol.

N(H) = 4.63·10²²; number of hydrogen atoms in sucrose.

N(O) = n(O) · Na.

N(O) = 0.0385 mol · 6.022·10²³ 1/mol.

N(O) = 2.31·10²²; number of oxygen atoms in sucrose.

Na is Avogadro constant.

4 0
3 years ago
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