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3 years ago

3.8x10^-4 g is dissolved into 450g of water . Find the concentration in ppm

Chemistry

1 answer:

IrinaK [193]3 years ago

4 0

Answer:

0.84ppm

Explanation:

Mass of solute = 3.8 x 10⁻⁴g

Mass of solution = 450g

Unknown is the concentration in ppm; parts per million

The parts per million is one of the units of representing small concentration of solutes in a solution.

Solutes are the substances dissolving in another to give solution.

Here the concentration is expressed in a million units.

Concentration in ppm = $\frac{mass of solute}{mass of solution}$ x 10⁶

= $\frac{3.8 x 10^{-4} }{450}$ x 10⁶

= 0.84ppm

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3 years ago

Helium gas diffuses 4 times as fast as an unknown gas. What is relative molecular mass of the gas

Inessa05 [86]

The relative molecular mass of the gas : 64 g/mol

<h3>Further explanation</h3>

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Required

The relative molecular mass of the gas

Solution

Graham's Law

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

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r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas

Input the value :

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3 years ago

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masya89 [10]

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4 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

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3 years ago