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mezya [45]
2 years ago
6

You have two solutions of chemical A. To determine which has the highest concentration of A (molarity), which of the following m

ust you know (there may be more than one answer)?a. the mass in grams of A in each solutionb. the molar mass of Ac. the volume of water added to each solutiond. the total volume of the solution
Chemistry
1 answer:
Sav [38]2 years ago
8 0

Answer:

a, b and c

Explanation:

To calculate molarity the number of moles and volume in dm3 is needed. So with the mass and molar mass the number of moles can be gotten and hence the molarity when dividing the number of moles by the volume in dm3.

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For the following reaction, 142 grams of silver nitrate are allowed to react with 22.3 grams of copper . silver nitrate(aq) copp
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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

and,

As, 3 mole of CaCl_2 react to give 1 mole of Ca_3(PO_4)_2

So, 0.0901 mole of CaCl_2 react to give \frac{1}{3}\times 0.0901=0.030 mole of Ca_3(PO_4)_2

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

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g

and,

\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2

\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g

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

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3 years ago
Find the value of 6 +<br>6 +<br>6 + 16 + vo​
AVprozaik [17]

Answer:

6

Explanation:

the value is 6 because its an even number

7 0
3 years ago
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