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sveticcg [70]
3 years ago
5

Please help me. I can’t give too much points bc I don’t have too much.

Chemistry
1 answer:
zubka84 [21]3 years ago
7 0
Answer is photons hope this helps
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N2 + 3 H2 → 2 NH3
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Answer:

B

Explanation:

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What are 3 permeable materials
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Permeable materials for what 
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The balanced equation for photosynthesis is shown below.
klio [65]
The true answer is: It's conserved because the total number of H atoms on each side is 12.
the first two answer is wrong because it's conserved not as mentioned, It's not conserved.

and the last one also wrong because the total number of O atoms are equal at the two sides but not equal 2.

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How many grams of benzene (C6H6) will produce 42 grams of CO2?
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The balanced chemical reaction is:<span>

</span><span>2C6H6 + 15O2 → 12CO2 + 6H2O</span><span>

We are given the amount of carbon dioxide to be produced for the reaction. This will be the starting point of our calculations.

</span>42 g CO2 ( 1 mol CO2 / 44.01 g CO2) ( 2 mol C6H6 / 12 mol CO2 ) (78.1074 g C6H6 / 1 mol C6H6) = 12.42 grams of C6H6
7 0
3 years ago
It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0?C to 24.3?C. What is the specific heat for
svet-max [94.6K]

Answer:

c = 0.377 J/g.°C

c = 0.2350 J/g.°C

J = 27.3 J

Explanation:

We can calculate the heat (Q) absorbed or released by a substance using the following expression.

Q = c × m × ΔT

where,

c: specific heat

m: mass

ΔT: change in the temperature

<em>It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0°C to 24.3°C. What is the specific heat for the metal? Express your answer numerically, in J/g.°C</em>

Q = c × m × ΔT

49.0 J = c × 11.5 g × (24.3°C - 13.0°C)

c = 0.377 J/g.°C

<em>The molar heat capacity of silver is 25.35 J/mol.°C. How much energy would it take to raise the temperature of 11.5g of silver by 10.1°C? Express your answer numerically, in Joules. What is the specific heat of silver?</em>

<em />

The molar mass of silver is 107.87 g/mol. The specific heat of silver is:

\frac{25.35J}{mol.\° C} .\frac{mol}{107.87g} =0.2350J/g.\° C

Q = c × m × ΔT

Q = (0.2350 J/g.°C) × 11.5 g × 10.1°C = 27.3 J

5 0
3 years ago
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