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Ahat [919]
4 years ago
15

What mass of iron is found in 280 g of 2 ppm solution?

Chemistry
1 answer:
Nataly_w [17]4 years ago
6 0

<u>Answer:</u> The mass of iron found in the solution is 0.56 mg

<u>Explanation:</u>

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

Both the masses are in grams.

We are given:

Concentration of iron = 2 ppm

Mass of solution = 280 g

Putting values in above equation, we get:

2=\frac{\text{Mass of iron}}{280g}\times 10^6\\\\\text{Mass of iron}=\frac{2\times 280}{10^6}=0.00056g=0.56mg

<u>Conversion factor used:</u>  1 g = 1000 mg

Hence, the mass of iron found in the solution is 0.56 mg

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A sample contains 6.73 g of Na2CO3 in water to make a total of 250.0 mL of solution. What is the molarity of sodium carbonate in
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Answer:

Molarity of Na₂CO₃ = 0.25M

% mass = 2.69

Explanation:

Molarity means mole of solute in 1L of solution

Molar mass of solute (Na₂CO₃) = 105,98 g/m

Moles = mass / molar mass → 6.73 g / 105.98 g/m = 0.0635 m

Mol/L = [M]

0.0635 mol/0.250L = 0.25M

Density of solution = Solution mass / Solution volume

1 g/ml = Solution mass / 250 mL → Solution mass is 250g

% mass will be:

In 250 g of solution we have 6.73 g of solute

in 100 g of solution we have (100 . 6.73)/250 = 2.69

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3 years ago
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Nail polish remover containing acetone was spilled in a room 7.2 m × 5.8 m × 3.9 m. Measurements indicated that 6,400 mg of acet
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Volume of room  = 7.2 m \times 5.8 m \times  3.9 m

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Now, according to conversion factor, convert 6,400 mg to micrograms

Since, 1 mg is equal to 1000 microgram.

Therefore, 6,400 mg  = 6,400 \times 1000

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To calculate concentration of acetone, divide volume and evaporated amount of acetone in micrograms.

Thus,

Concentration of acetone  = \frac{6,400,000 microgram}{162.864 m^{3}}

= 39296.5910 microgram per cubic meter or 3.92\times 10^{4} \mu g/m^{3}\simeq 4.0\times 10^{4} \mu g/m^{3}

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3 years ago
Thisss for you blake :)))
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I’m not Blake but ok
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Consider the following intermediate chemical equations.
mixas84 [53]

Answer:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

Explanation:

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CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

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<em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).</em>

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Which scientist is credited with developing the first scientific atomic theory
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