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Vanyuwa [196]
2 years ago
11

A 98.0°C piece of cadmium (c=.850J/g°C) is placed in 150.0g of 37.0°C water. After sitting for a few minutes, both have a temper

ature of 38.6°C. What was the mass of the cadmium sample?
Chemistry
1 answer:
Eduardwww [97]2 years ago
3 0
The answer is 19.9 grams cadmium. 
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
     heat lost by cadmium = heat gained by the water
     -qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
     -qcadmium = qwater
     -(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
     mcadmium = 19.9 grams
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1N2 + 3H2 -->
Hunter-Best [27]

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

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2 years ago
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Answer:

B. Corn needs rain to grow, but too much rain is harmful.

7 0
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These substances are all ____. H2O2 Li NaCl O2 Elements Pure Substances Molecules Compounds
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Your answer is Compounds

6 0
3 years ago
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Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if 2 moles of pentane (C5H12) were supplied, and an unlimited amount
Kamila [148]

12 moles of water H₂O are produced from the combustion of pentane.

Explanation:

We have the following combustion of pentane (C₅H₁₂):

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Knowing the chemical reaction we devise the following reasoning:

if         1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O

then    2 moles of pentane C₅H₁₂ produces X moles of water H₂O

X = (2 × 6) / 1 = 12 moles of water H₂O

Learn more about:

combustion of organic compounds

brainly.com/question/7295137

brainly.com/question/884053

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Answer:

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