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marishachu [46]

3 years ago

8

what occurs in the chemical reaction (A) reactions break bonds and reform the same bonds (B) no chemical bonds are broken (C) re

actants break bonds and form new bonds to make new substance (D) reactants break bonds but do not form new substances

1 answer:

Likurg_2 [28]3 years ago

4 0

The answer is Cc reactants break bonds and form new bonds to make new substances

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Sarah rolls a ball in the grass. The ball starts off moving quickly, starts to slow down, and eventually stops. Using the words

gogolik [260]

Answer:

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Explanation:

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3 years ago

What is the boiling point of water when 175.0 g of Na2SO4, a strong electrolyte is dissolved in 1.000 Kg of water?

liubo4ka [24]

Answer: $101.9^0C$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0=(T_b-100)^0C$ = Elevation in boiling point

i= vant hoff factor = 3 (number of ions an electrolyte produce on complete dissociation)

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

$K_f$ = freezing point constant = $0.512^0C/m$

m= molality

$\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (water)= 1.000 kg

Molar mass of solute $Na_2SO_4$ = 142 g/mol

Mass of solute $Na_2SO_4$ = 175.0 g

$(T_b-100)^0C=3\times 0.512\times \frac{175.0g}{142g/mol\times 1.000kg}$

$T_b=101.9^0C$

Thus the boiling point of water when 175.0 g of $Na_2SO_4$ , a strong electrolyte is dissolved in 1.000 Kg of water is $101.9^0C$

8 0

3 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{E}{E'}=\sqrt{\frac{M'}{M}}$

$\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}$

M' = 70.77 g/mol

70.77 g/mol is the molar mass of the unknown gas.

8 0

3 years ago