Formula for the compound that contains Mg2+ and O2-

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.

5 0

3 years ago

A large highway barrier is 1 meter wide, by 1 meter tall, by 2 meters long.

Romashka [77]

Answer:

when mass is 1×10⁴ Kg then density is 5 g/cm³.

when mass is 104 Kg then density is 5.2 × 10⁻² g/ cm³.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

mass = 1×10⁴ Kg

volume= w ×l× h = 1×2× 1 = 2 m³

density = ?

first of all we will convert the given volume meter cube to cm³:

we know that

2×1000000 = 2 × 10⁶ cm³

Now we will convert the mass into gram.

1 Kg = 1000 g

1×10⁴ × 1000 = 1 ×10⁷ g

Now we will put the values in the formula,

d = m/v

d = 1 ×10⁷ g / 2×10⁶ cm³

d = 0.5 × 10¹ g/cm³

or

d = 5 g/cm³

If mas is 104 Kg:

104 × 1000 = 104000 g

d= m/v

d = 104000 g / 2×10⁶ cm³

d= 52000 ×10⁻⁶ g/ cm³

d= 5.2 × 10⁻² g/ cm³

7 0

3 years ago

Which causes waves in the ocean to occur?

galina1969 [7]

Answer:

Explanation:

C. energy being transferred from wind

Waves are most commonly caused by wind. Wind-driven waves, or surface waves, are created by the friction between wind and surface water. As wind blows across the surface of the ocean or a lake, the continual disturbance creates a wave crest. The gravitational pull of the sun and moon on the earth also causes waves.

4 0

3 years ago

Name the following compounds CaSe <br><br> don’t use common names such as water or salt

Contact [7]

Do it urself ;) jdhdhdhdhdhhdhdhdhdhdhdhdhdhgdhd

8 0

3 years ago

When is a roman numeral most likely needed in the name of an ionic compound?

Sauron [17]

The answer is I’m guessing c

5 0

3 years ago

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