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3 years ago

The molarity of a 4.200 L solution is 1.230 M Na2CO3. What is the mass of Na2CO3

Chemistry

1 answer:

Lerok [7]3 years ago

7 0

Answer:

547.5g

Explanation:

To get the mass, you need moles.

moles = (molarity)(Liters)

moles = (1.230M)(4.200L) = 5.166 moles Na2CO3

Now, just use stoichiometry

molar mass of Na2CO3 = 2(mass of Na) + (mass of C) + 3(mass of O)

= 2(22.9) + 12.01 + 3(16) = 105.99g/mol

5.166moles(105.99g/mol)

= 547.544

But, the measurements given had 4 significant figures, so in chemistry we write:

547.5g

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The rate of geologic processes can happen quickly or take a long time.

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Answer:

Take a long time

Explanation:

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3 years ago

What is true when a reaction has reached equilibrium?

goldfiish [28.3K]

Answer:

In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. A catalyst speeds up the rate of a chemical reaction, but has no effect upon the equilibrium position for that reaction.

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3 years ago

Read 2 more answers

Determine the electron-group arrangement, molecular shape, and ideal bond angle(s) for each of the following:

lubasha [3.4K]

Solution :

a). $SO_3$

This compound is known as sulfur trioxide.

The molecular shape of sulfur trioxide is trigonal planer.

And the bond angle is 120°.

b). $N_2O$

This compound is known as Nitrous oxide. Here, nitrogen is in the center. There is no lone pair around the nitrogen atom and it forms two sigma bonds with the other two atoms.

It is linear in shape.

The bond angle between them is 180°.

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3 years ago

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of

Ira Lisetskai [31]

Answer:

a. 167 mL.

b. 39.3 %.

Explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles as shown below:

$n_{H_2SO_4}=45.0gAl*\frac{1molAl}{27.0gAl} *\frac{3molH_2SO_4}{2molAl} =2.50molH_2SO_4$

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

$V=\frac{n}{M}=\frac{2.50mol}{15.0mol/L}=0.167L$

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

$m_{Al_2(SO_4)_3}=2.50molH_2SO_4*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{342.15gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=285.13gAl_2(SO_4)_3$

Therefore the percent yield is:

$Y=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%$

Best regards!

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2 years ago

Te second period at the bottom of the table is called

katovenus [111]

Answer:

The second row elements are called Actinidies . abbreviated as Ac

Explanation:

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3 years ago