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3 years ago

A 40 g gold coin is cooled from 50°C to 10°C (CAu is 0.13 J/g-°

1 answer:

4 0

When the Heat gain or lose = the mass * specific heat * ΔT

and when we have the mass of gold coin= 40 g

and the specific Heat of gold= 0.13 J/g°

and ΔT = (Tf- Ti) = 10°C - 50°C = -40 °C

so by substitution:

∴Heat H = 40 g * 0.13 J/g° * -40
= - 208 J

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The specific heat capacity of gold is 0.128 J/g C. How much would be needed to warm 250.0 grams of gold from 25.0 C to 100.0 C.

madam [21]

Answer:

3,200 joules

Explanation:

q = mcΔT = (250.0 g)(0.128 j/g°C)(100°C -25°C) = 3,200 joules

7 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

Answer: The limiting reactant is ammonia $(NH_3)$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For ammonia:

Given mass of ammonia = 135.9 kg = 135900 g (Conversion factor: 1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol$

For carbon dioxide gas:

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol$

The given chemical reaction follows:

$2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = $\frac{1}{2}\times 7994.12=3997.06mol$ of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia $(NH_3)$

5 0

3 years ago

The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O

Mila [183]

Answer:

41.3kJ of heat is absorbed

Explanation:

Based in the reaction:

Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ

1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.

63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:

63.4g Fe₃O₄ × (1mol / 231.533g) = 0.274moles of Fe₃O₄

These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:

0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =

<h3>41.3kJ of heat is absorbed</h3>

6 0

3 years ago

What amount of heat is released when the temperature of 450.0 g of a

Molodets [167]

Answer:

$\boxed {\boxed {\sf 4038.48 \ Joules}}$

Explanation:

Since we are given the mass, specific heat, and change in temperature, we should use this formula for heat:

$q=mc\Delta T$

The substance's mass is 450.0 grams, the specific heat is 1.264 J/g°C, and the change in temperature is 7.1 °C.

$m= 450.0 \ g \\c= 1.264 \ J/g \textdegree C\\\Delta T= 7.1 \ \textdegree C$

Substitute the values into the formula.

$q= (450.0 \ g)(1.264 \ J/g \textdegree C)(7.1 \ \textdegree C)$

Multiply the first 2 values together. The grams will cancel out.

$q= 568.8 \ J/ \textdegree C (7.1 \ \textdegree C)$

Multiply again. This time, the degrees Celsius cancel out.

$q= 4038.48 \ J$

4038.48 Joules of heat energy are released.

5 0

3 years ago

If 48 g of magnesium reacts with oxygen gas, how many grams of magnesium oxide will be formed according to the following equatio

Aloiza [94]

think it maybe 48 mg0 because there should be 2g magnesium in the word equation

4 0

2 years ago

Read 2 more answers