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grin007 [14]
3 years ago
5

You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution

is known. The concentration of the KCN solution given in Part A corresponds to a mass percent of 0.480 %. What mass of a 0.480 % KCN solution contains 697 mg of KCN? Express the mass to three significant figures and include the appropriate units. View Available Hint(s)
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

145 grams of a 0.480 % KCN solution contains 697 mg of KCN.

Explanation:

w/w % : The percentage mass or fraction is mass of the of solute present in 100 grams of the solution.

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

given =  w/w% = 0.480 %

Mass of solute that KCN = 697 mg = 0.697 g

(1 mg = 0.001 g)

Mass of the solution = ? = x

0.480\%=\frac{0.697 g}{x}\times 100

x=\frac{0.697 g}{0.480}\times 100

x = 145.21 g ≈ 145 g

145 grams of a 0.480 % KCN solution contains 697 mg of KCN.

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V∝ T, P  is constant  

where V, T and P are volume, temperature and pressure

\frac{V1}{T1 } = \frac{V2}{T2}

where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.  

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3 years ago
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
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Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

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(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

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Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

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Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

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Answer:

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Sign convention is, w = (-ve)

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