Answer : Note : The formula give in the question needs to be corrected , the formula of two chlorine atoms for every magnesium atom in the compound is $MgCl_{2}$ which is Magnesium Chloride.

Explanation : To write the formula of the compound we need to know the charge of the atoms or elements involved in the formula.

Charge of Mg is +2 and charge of Cl is -1 , to write the formula of the compound we always criss cross the charges. And also when we write the formula the total (or net) charge of the compound should be 0. Charge of Mg is +2 so to get the total(net) charge as 0 we need two Cl atoms with each -1 charge.

Mg = +2 , Cl = -1 , Cl = -1

2 Cl = -1 + (-1) = -2

Mg = +2 and 2 Cl = -2

+2 -2 = 0

$MgCl_{2}\rightarrow Mg^{+2} + 2Cl^{-}$

So the formula for Magnesium Chloride is $MgCl_{2}$

8 0

3 years ago

Please help, How could you tell a Ca(NO3)2 solution from a Zn(NO3)2 solution?

tensa zangetsu [6.8K]

Answer:

Anhydrous calcium nitrate reaction with water is highly endothermic

8 0

1 year ago

Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 wa

Alex777 [14]

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

A sample contains 19.3 g of $N_2$ is reacted with 19.3 g of $F_2$ . Now we need to find the amount of $NF_3$ that can be formed by the complete reactions of each of the reactants.

If all of the $N_2$ was used up in the reaction, how many moles of $NF_3$ would be produced?

Answer: 1.378 moles of $NF_3$ are produced in the reaction.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, $N_2$ is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of $N_2$ = 19.3 g

Molar mass of $N_2$ = 28.02 g/mol

Putting values in equation 1:

$\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol$

For the given chemical reaction:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

By the stoichiometry of the reaction:

1 mole of $N_2$ produces 2 moles of $NF_3$

So, 0.689 moles of $N_2$ will produce = $\frac{2}{1}\times 0.689=1.378mol$ of $NF_3$

Hence, 1.378 moles of $NF_3$ are produced in the reaction.

8 0

3 years ago

Which substance is an element? A. Methane (CH4) B. Baking Soda aka Sodium Bicarbonate (NaHCO3) C. Einsteinium (Es) D. Bleach aka

zloy xaker [14]

Answer:

Explanation:

4 0

4 years ago