Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

Chemistry

1 answer:

Minchanka [31]2 years ago

6 0

Answer:

Mn(NO3)3 Molar mass = (54.93) + (3 x 14) + (9 x 16) = 240.93g/mol

Total number of Oxygen atoms in formula = 9

Total mass of oxygen atoms theoretically in the formula = 9x16 = 144g/mol

Oxygen's percentage composition:

(144/240.93) x 100 = 59.7% of the compound is oxygen

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Al2(SO4)3+Mg(NO3)2⟶ What would be the product(s) of this reaction? *These are NOT balanced, just look for the correct products*

Readme [11.4K]

Answer:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Explanation:

Hello there!

In this case, for the given reactants side, we infer this is a double replacement reaction because all the cations and anions are switched around as a result of the chemical change, we infer that the products side include aluminum with nitrate and magnesium with sulfate as shown below:

$Al_2(SO_4)_3+Mg(NO_3)_2\rightarrow Al(NO_3)_3+MgSO_4$

However, we need to balance since unequal number of atoms are present at both sides, thus, we do that as shown below:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Thus, we make 6 Al atoms, 3 S atoms, 3 Mg atoms and 30 O atoms on each side in agreement with the law of conservation of mass.

Regards!

7 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$