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Rudiy27
3 years ago
9

Find the mass in grams of 2.00 * 10 ^ 23 molecules of dinitrogen pentoxide gas.

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

idk srr :(

Explanation:

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Water is always subject to the law of
dimulka [17.4K]

Answer:

motion

Explanation:

i just know things

5 0
3 years ago
Essentially, the oxygen we breath is
stepan [7]

Answer:

b

Explanation:

Plants need carbon dioxide and let out oxygen

3 0
3 years ago
A sample of 3.62 moles of diphosphorous trioxide is
Sindrei [870]

H_3PO_32H_3PO_32H_3PO_3P_2O_3Answer:

B

Explanation:

This question is about stoichiometry. From the balanced equation P_2O_3 + 3H_2O⇒2H_3PO_3, we see that 3 moles of water is needed to react with 1 mole of P_2O_3.

This means that, to fully react 3.62 moles of P_2O_3, we would need 3*3.62 or  10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of P_2O_3, 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of P_2O_3.

From the balanced equation, we see that every mole of P_2O_3 reacted gets you 2 moles of 2H_3PO_3. Therefore, 2.1033 moles of P_2O_3 would give you approximately 4.21 moles of H_3PO_3.

5 0
2 years ago
When air pressure rapidly falls, what weather change usually occurs?
Mama L [17]

hope this helps

Explanation:

a prolong storm will occur

that's what I found

8 0
3 years ago
Calculate the heat absorbed by the water in a calorimeter when 175 grams of lead cools from 125.0°C to 22.0°C. The specific heat
Iteru [2.4K]

Answer:

Q = 233.42 J

Explanation:

Given data:

Mass of lead = 175 g

Initial temperature = 125.0°C

Final temperature = 22.0°C

Specific heat capacity of lead = 0.01295 J/g.°C

Heat absorbed by water = ?

Solution:

Heat  absorbed by water is actually the heat lost by the metal.

Thus, we will calculate the heat lost by metal.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 22.0°C - 125.0°C

ΔT = -103°C

Q = 175 g × 0.01295 J/g.°C×-103°C

Q = -233.42 J

Heat absorbed by the water is 233.42 J.

5 0
3 years ago
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