Ask question

Ask question

All categories

3 years ago

13

Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3

) 2 ) . Round your answer to significant digits.

1 answer:

Pachacha [2.7K]3 years ago

6 0

<u>Answer:</u> The volume of barium chlorate is 195.65 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

$0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL$

Hence, the volume of barium chlorate is 195.65 mL

You might be interested in

Which of the elements shown will not form ions, and why will they not do so?

grigory [225]

No elements visible!

Ions form between metals and non-metals.

Hope this helps!

6 0

2 years ago

what is the mass of alumunium will be completly oxidized by 44.8 L of oxygen to produce Al2o3 at STP?

Kobotan [32]

Answer:

Therefore, 2 moles of oxygen will be required to oxidize 2.66 moles of oxygen. 72g of Aluminium will be completely oxidized by 44.8 lit of oxygen at STP.

3 0

3 years ago

How many carbon atoms are there in 1.00 mol of C2H6?

Sphinxa [80]

Since it's C2, then it must be two Carbon atoms. Think of it this way; there are two atoms of Hydrogen in a water molecule, so we now have H2O.

5 0

4 years ago

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

<u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

<u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

4 0

3 years ago

Which element has a larger atomic radius than sulfur?

Novay_Z [31]

Cadium

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period.

6 0

3 years ago

Read 2 more answers