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andrew11 [14]
3 years ago
11

A sample of 450.0 g of water is found to contain 180 mg Cd. What is this concentration in parts per million?

Chemistry
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

The correct answer is 399.8 ppm

Explanation:

A concentration in parts per million (ppm) is equal to:

1 ppm = \frac{1 g solute}{10^{6}g solution}= \frac{1 mg solute}{1 kg solution}

Solute: Cd; Mass = 180 mg x (1 g/1000 mg) = 0.18 g

Solvent: Water ; Mass= 450.0 g x (1 kg/1000 g) = 0.45 kg

We have the following total mass of solution:

Mass of solution = Mass of solute + Mass of solvent = 0.18 g + 450 g = 450.18 g = 0.45018 kg

Finally, we divide the <u>mass of solute (in mg)</u> into the <u>mass of solution (in kg)</u> to obtain the ppm (in mg/Kg):

ppm = 180 mg/0.45018 kg = 399.8 mg/Kg = 399.8 ppm

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If 156.06 g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 → CO2 + H2O Select one:
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Answer:

The correct option is;

a. 255.0 g

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The equation of the combustion reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

From the balanced chemical equation of the reaction, we have;

One mole of propane, C₃H₈ reacts with five moles oxygen gas, O₂, to form three moles of carbon dioxide, CO₂, and  four moles of water, H₂O

The molar mass of propane gas = 44.1 g/mol

The number of moles, n, of propane gas = Mass of propane gas/(Molar mass of propane gas) = 156.06/44.1 = 3.54 moles

Given that one mole of propane gas produces 4 moles of water molecule (steam) H₂O, 3.54 moles of propane gas will produce 4×3.54 = 14.16 moles of  (steam) H₂O

The mass of one mole of H₂O = 18.01528 g/mol

The mass of 14.16 moles of H₂O = 14.16 × 18.01528 = 255.0 g

The mass of H₂O produced = 255.0 g

3 0
3 years ago
What is the solubility of MgCO₃ in a solution that contains 0.080 M Mg²⁺ ions? (Ksp of MgCO₃ is 3.5 × 10⁻⁸)
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Answer: 4.4 x 10^-7

Explanation:

The dissociation equation for this reaction is:

MgCO3 (s) → Mg+2 (aq) + CO3-2 (aq)

\text { So, } k_{s p}=(x+0.08) x (Here 0.08 >>> x )

\begin{aligned}\Rightarrow 3.5 \times 10^{-8} &=0.08 \times x \\\Rightarrow x &=\frac{3.5}{0.08} \times 10^{-8} \\&=4.37 \times 10^{-7} \\\Rightarrow x & \approx 4.4 \times 10^{-7} \mathrm{~M}\end{aligned}

So the solubility MgCO₃ in a solution that containing 0.080 M Mg²⁺ is 4.4 x 10^-7

7 0
2 years ago
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