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kkurt [141]
3 years ago
7

A student is told to use 20.0 grams of sodium chloride to make an aqueous solution that has a concentration of 10.0 grams of sod

ium chloride per liter of solution. assuming that 20.0 grams of sodium chloride has a volume of 7.5 milliliters, about how much water will she use in making this solution?
Chemistry
1 answer:
Butoxors [25]3 years ago
7 0

Volume of the solution = 20.0 g NaCl * \frac{1 L solution}{10.0 g NaCl}

= 2 L solution x \frac{1000 mL}{1 L} =2000 mL

Volume of solute = 7.5 mL

Volume of water (solvent) = 2000 mL - 7.5 mL = 1992.5 mL water

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A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
(1) (a) In an equilibrium reaction between gases Q2 and R2 to form QR, the energy content of the reactants is 100KJ and that of
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Answer:

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Explanation:

7 0
2 years ago
Al(OH)3 + 3 HCl = AlCl3 + 3 H2O
Kryger [21]

Answer:

1) 1.235 g.

2) 0.61 g.

Explanation:

  • From the balanced equation:

<em>Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O.</em>

1.0 mol of Al(OH)₃ reacts with 3.0 moles of HCl to produce 1.0 mol of AlCl₃ and 3.0 moles of H₂O.

<em>1) How many grams of HCl can a tablet with 0.880 g of Al(OH)₃ consume? </em>

  • To calculate the amount of HCl needed to consume 0.880 g of Al(OH)₃, we need to calculate the no. of moles of Al(OH)₃:

no. of moles of Al(OH)₃ = mass/molar mass = (0.880 g)/(78.0 g/mol) = 1.13 x 10⁻² mol.

∵ Every 1.0 mol of Al(OH)₃ needs 3.0 moles of HCl to be consumed.

∴ 1.13 x 10⁻² mol of Al(OH)₃ needs (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of HCl.

The no. of grams of HCl = no. of moles of HCl x molar mass of HCl = (3.385 x 10⁻² mol)(36.5 g/mol) = 1.235 g.

<em>2) How much H₂O?</em>

∵ Every 1.0 mol of Al(OH)₃ produces 3.0 moles of H₂O.

∴ 1.13 x 10⁻² mol of Al(OH)₃ produces (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of H₂O.

<em>The no. of grams of H₂O = no. of moles of H₂O x molar mass of H₂O </em>= (3.385 x 10⁻² mol)(18.0 g/mol) = <em>0.6092 g ≅ 0.61 g.</em>

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