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2 years ago

What is the mass of 3.2 moles of Sulfur?

Chemistry

1 answer:

FinnZ [79.3K]2 years ago

8 0

Answer:

820.864 g

Explanation:

1) The atomic mass of sulfur (found from the periodic table) is 32.065 amu. Use this mass to find the molar mass of Sulfur. Sulfur is S8 so the molar mass of sulfur is:

8 × 32.065 = 256.52 g/mol

2) To find the mass use the formula:

m = n × M where m is the mass, n is the number of moles, and M is the molar mass.

$m \: = 3.2 \times 256.52$

$m = 820.864 \: g$

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3 years ago

Read 2 more answers

Boyle's law only works when a gas is kept at a constant temperature. Experimentally this is very tricky as changes in pressure o

Lady bird [3.3K]

The heat that creates this temperature change coming from change in the internal energy of the system as per as first law of thermodynamics.

<h3>What is Boyle's law ?</h3>

A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

As we know, Boyle's law only works when the gas is kept at a constant temperature

Here,

When volume of gases decreased, it means work done has occurred on the system, so the work done is used for raising internal energy of the gas and the other is released as the thermal energy.

So,

According to 1st law of thermodynamics,

we know Q = ΔU + W i.e, change in internal energy and work done. So this is a reason. Changing temperature occurs.

Learn more about Internal enrgy here ;

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2 years ago

2 HCl + 1 Zn -> 1 H2 + ZnCl2 what type of reaction is this

Salsk061 [2.6K]

Answer:

Single replacement reaction

Explanation:

3 0

3 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 $HCl$ .

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