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Scorpion4ik [409]
3 years ago
6

A 716.4g sample of iron is subjected to a temperature change of -33.47°C while releasing 996.6 joules of heat. What is the speci

fic heat capacity of iron?
Answer choices:
29.8 J/g°C
0.04 J/g°C
-0.04 J/g°C
1.39 J/g°C
Chemistry
1 answer:
kenny6666 [7]3 years ago
7 0

Answer:

Specific heat capacity of iron (c) = -0.04 J/g°C

Explanation:

Given:

Mass of sample of iron M = 716.4 gram

Change in temperature Δt = -33.47°C

Heat energy Q = 996.6 joules

Find:

Specific heat capacity of iron (c)

Computation:

We know that

Q = McΔt

So,

c = Q / M[Δt]

Specific heat capacity of iron (c) = 996.6 / (716.4)(-33.47)

Specific heat capacity of iron (c) = 996.6 / -23,977.908

Specific heat capacity of iron (c) = -0.0415

Specific heat capacity of iron (c) = -0.04 J/g°C

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4 0
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The solubility of copper chloride at 20 °C is 73 g/100 g of water. Kiera adds 100 g of copper chloride to 100 g of water and sti
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Answer:

m_{undissolved}=27g

Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

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Best regards!

3 0
2 years ago
Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react
Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of BCl_3 = 60 g

Mass of H_2O = 37.5 g

Molar mass of BCl_3 = 117 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of HCl = 36.5 g/mole

First we have to calculate the moles of BCl_3 and H_2O.

\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles

\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)

From the balanced reaction we conclude that

As, 1 mole of BCl_3 react with 3 mole of H_2O

So, 0.513 moles of BCl_3 react with 3\times 0.513=1.539 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and BCl_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of HCl.

As, 1 mole of BCl_3 react to give 3 moles of HCl

So, 0.513 moles of BCl_3 react to give 3\times 0.513=1.539 moles of HCl

Now we have to calculate the mass of HCl.

\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl

\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0
3 years ago
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