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4 years ago

Question 1.

Chemistry

1 answer:

Travka [436]4 years ago

5 0

Answer:

Question 1. → The amount remained is 15.625 g after 5 half-lives (30 hours).

Question 2. → mass % of water =41.4 %

Question 3. → C. 64 grams.

Explanation:

Question 1. 

It is known that the decay of isotopes and radioactive material obeys first order kinetics.

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

Thus, 30 hours represent (30/6) 5 half-lives.

500 g →(first 1/2 life) 250 g →(second 1/2 life) 125 g →(third 1/2 life) 62.5 g →(fourth 1/2 life) 31.25 g →(fifth 1/2 life) 15.625 g

So, The amount remained is 15.625 g after 5 half-lives (30 hours).

Question 2.

The formula of Copper (II) Fluoride tetra hydrate is CuF₂.4H₂O.

the molar mass of Copper (II) Fluoride tetra hydrate is the sum of the atomic masses o each element.

molar mass of 1 mol of CuF₂.4H₂O = 64 + 2*(19) + 8* (1) + 4*(16) = 174 g/mol.

molar mass of 4 mol of H₂O = 8*(1) + 4*(16) = 72 g/mol

1 mol of Copper (II) Fluoride tetra hydrate → 4 mol H₂O

mass % of water = (mass of 4 mol of H₂O) / (mass of 1 mol of CuF₂.4H₂O ) *100 = (72 /174) *100 = 41.4 %

Question 3.

Mass percentage is defined as the mass of a solute divided by the total mass of the solution, multiplied by 100

mass percent = ( mass of solute / mass of solution) *100

∴ 16 = ( mass of solute / 400 g) *100

mass of solute = (16* 400)/ 100 = 64 g.

So the right choice is C. 64 grams.

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4 years ago

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

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Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

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If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

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$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

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4 years ago

A small amount of a solid is added to water. The observation made after fifteen minutes is shown in the figure. Which of these s

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Answer:

B) sand

Explanation:

A) Oil is a wrong choice because it is a liquid not a solid and also if it is oil, it will float over the water surface as a droplets.

B) Sand is the right choice, because sand is a solid and it does not dissolve in water and stabilizes at the bottom.

C) Sugar is a wrong choice, because small amount of sugar will dissolve in water and be a homogeneous solution and does not appear as a particles.

D) Wood ships is also a wrong choice, even it is a solid and does not dissolve in water, but it will float over the water surface.

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4 years ago

Calculate the number of grams in 4.56 x 1026 atoms of sodium phosphate. Be sure to balance the charges of sodium phosphate. Help

zalisa [80]

Answer:

124225.91 g of Na₃PO₄

Explanation:

From the question given above, the following data were obtained:

Number of atoms of Na₃PO₄ = 4.56×10²⁶ atoms

Mass of Na₃PO₄ =?

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Na₃PO₄

Next, we shall determine the mass of 1 mole of Na₃PO₄. This can be obtained as follow:

1 mole of Na₃PO₄ = (23×3) + 31 + (16×4)

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Thus,

6.02×10²³ atoms = 164 g of Na₃PO₄

Finally, we shall determine the mass of Na₃PO₄ that contains 4.56×10²⁶ atoms. This can be obtained as follow:

6.02×10²³ atoms = 164 g of Na₃PO₄

Therefore,

4.56×10²⁶ atoms = (4.56×10²⁶ × 164)/6.02×10²³

4.56×10²⁶ atoms = 124225.91 g of Na₃PO₄

Therefore, 124225.91 g of Na₃PO₄ contains 4.56×10²⁶ atoms

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3 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

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

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

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(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

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