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3 years ago

_Ba3Na2+ H2O + Ba(OH)2 + NH3 Balance this equation

Chemistry

1 answer:

kati45 [8]3 years ago

6 0

Answer:

Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.

Ionic charges are not yet supported and will be ignored.

Replace immutable groups in compounds to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will.

Compound states [like (s) (aq) or (g)] are not required.

You can use parenthesis () or brackets [].

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It is the expression of Charles' Law.

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According to this law, the volume of a fixed mass of a gas is directly proportional to its temperature and is given by:

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

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3 years ago