What causes a change in a rocks compostion

Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -> 1 ______ +2LiF

ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0

2 years ago

Question 11 (10 points)

andrey2020 [161]

Answer:

The correct answer is option a.

Explanation:

When aluminum hydroxide reacts with of nitrous acid it gives of aluminum nitrite and of water.

$Al(OH)_3+3HNO_2\rightarrow Al(NO_2)_3+3H_2O$

According to above reaction ,when 1 mole of aluminum hydroxide reacts with 3 moles of nitrous acid it gives 1 mole of aluminum nitrite and 3 moles of water.

Hence, the correct answer is option a.

4 0

3 years ago

With no difference in temperatures and 100% humidity, will it rain? Why or Why not?

Artyom0805 [142]

Not necessarily, rain needs some mechanism such as instability of vertical air movement..

7 0

3 years ago

An organic compound, which has the empirical formula C12H25 has an approximate molar mass of 338 g/mol. What is its probable mol

Marta_Voda [28]

Answer:

C24H50

Explanation:

The empirical fomula's molar mass is 169.25 g/mol.

We know the molecular formula's molar mass is 338 g/mol.

338/169.25= 1.99 or approximately 2

8 0

3 years ago

A solid powder is known to be a mixture of NaCl and Na2CO3, but the relative amounts of each compound in the sample are unknown.

AveGali [126]

Answer:

Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.

Explanation:

$Na_2CO_3(aq) +2 HCl(aq)\rightarrow 2NaCl(aq) + H_2O(l)+CO_2(g)$

Molarity of HCl solution = 0.1174 M

Volume of HCl solution = 83.15 mL = 0.08315 L

Moles of HCl = n

$molarity=\frac{moles}{Volume (L)}$

$0.1174 M=\frac{n}{0.08315 L}$

$n=0.1174 M\times 0.08315 L=0.009762 mol$

According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then 0.009762 mol of HCl will recat with:

$\frac{1}{2}\times 0.009762 mol=0.004881 mol$

Moles of Sodium carbonate = 0.004881 mol

Volume of the sodium carbonate containing solution taken = 1L

Concentration of sodium carbonate in the solution before the addition of HCl:

$[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L$

4 0

3 years ago