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2 years ago

What is the amount of heat required to complete MELT a 200 gram sample of H2O(s)

Chemistry

2 answers:

wolverine [178]2 years ago

8 0

Answer:

66800 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 g

Heat (Q) required =?

The heat required to melt the 200 g sample of water can be obtained as:

Mass (m) = 200 g

Latent heat of water (L) = 334 J/g

Heat (Q) required =?

Q = mL

Q = 200 × 334

Q = 66800 J

Thus, 66800 J of heat energy is required to melt the sample of water.

alisha [4.7K]2 years ago

7 0

Answer:

66800 J

Explanation:

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3 years ago

Read 2 more answers

Write Balance chemical reaction for preparation of chlorine with or without application heat<br>

suter [353]

Answer:

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

Explanation:

Chlorine is a diatomic halogen gas known for its greenish-yellow colour. It has a pungent smell and is only moderately soluble in water.

It is a very reactive gas and is never found in free state in nature.

Chlorine can be prepared in the laboratory by oxidation of hydrochloric acid using KMnO4 as follows;

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

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3 years ago

PLS HELP What is the mass of hydrogen atoms that is measured at 54 u?

tankabanditka [31]

The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g

<h3>Conversion scale </h3>

1 u = 1.66×10¯²⁴ g

<h3>How to determine the mass of hydrogen atoms </h3>

Mass of Hydrogen (u) = 54 u
Mass of Hydrogen (g) =?

1 u = 1.66×10¯²⁴ g

Therefore

54 u = 54 × 1.66×10¯²⁴ g

54 u = 89.64×10¯²⁴ g

Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g

Learn more about conversion:

brainly.com/question/2139943

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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

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3 years ago

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Answer:

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2 years ago