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mixer [17]
3 years ago
10

What are 4 energy transfer

Chemistry
1 answer:
Klio2033 [76]3 years ago
5 0
What kind of energy transfer
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What are the 4 variables that describe a gas
Phantasy [73]
Pressure, volume, temperature, # moles Pressure, volume and temperature, and moles of gas


Hope that helps!!!!
8 0
3 years ago
How many particles are in 1.40 x 10^7 mol of aluminum (Al)?
SpyIntel [72]

Answer:

<h3>Hlo there !! </h3>

<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.</u>

<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.So 1.04*107 mol of Al contains 1.40*107 * 6.022*1023 = 8.43*1030 structural units (in case of Al – atoms).</u>

<h3><u>8.43*1030 particles Al.</u></h3>

Explanation:

<h3>Hope this helps !!</h3>
5 0
2 years ago
How are lead and plastic foam different?
Natalka [10]
Lead is heavier and more resistant. Plastic foam is light and can easily be broken
 
4 0
3 years ago
Read 2 more answers
How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?
PIT_PIT [208]
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
7 0
3 years ago
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at
lana [24]

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

6 0
2 years ago
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