All categories

3 years ago

How many grams is needed to make the following solution?

Chemistry

1 answer:

Lisa [10]3 years ago

3 0

Answer:

Explanation:

yxyyyxyxxyxxy yy y gg ff ff d d f t r rr rr rr rr r r t tt t

You might be interested in

What volume of copper metal is needed to react with 3.5 moles of silver nitrate

leonid [27]

Answer:

3.40 g

Explanation: don't have one :P

6 0

2 years ago

Chemical Reactions

irakobra [83]

Answer:

It works by separating the fuel from the oxygen. The oxygen comes from the air. It is the same oxygen we breathe. Since the oxygen has to be in contact with the fuel, if you can coat the fuel with something that keeps the oxygen away, the fire will go out which is what the fire extinguisher does.

6 0

2 years ago

Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 to n = 3

tankabanditka [31]

5 0

3 years ago

Which type of reaction produces the most dangerous radioactive waste?

k0ka [10]

The correct answer to the question above is fussion of uranium. The fussion of uranium is the only type of reaction in which produces the most dangerous radioactive waste. The fusion of uranium means the binding of their atoms and produce a radioactive waste.

7 0

3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$