What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M
solution?

Answers

Answer 1

Answer:

Approximately [tex]19.1\; \rm g[/tex].

Explanation:

Number of moles of formula units of magnesium sulfate required to make the solution

The unit of concentration in this question is "[tex]\rm M[/tex]". That's equivalent to "[tex]\rm mol \cdot L^{-1}[/tex]" (moles per liter.) In other words:

[tex]c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}[/tex].

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

[tex]\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L[/tex].

Calculate the number of moles of [tex]\rm MgSO_4[/tex] formula units required to make this solution:

[tex]\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}[/tex].

Mass of magnesium sulfate in the solution

Look up the relative atomic mass data of [tex]\rm Mg[/tex], [tex]\rm S[/tex], and [tex]\rm O[/tex] on a modern periodic table:

[tex]\rm Mg[/tex]: [tex]24.305[/tex].[tex]\rm S[/tex]: [tex]\rm 32.06[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].

Calculate the formula mass of [tex]\rm MgSO_4[/tex] using these values:

[tex]M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}[/tex].

Using this formula mass, calculate the mass of that (approximately) [tex]0.159\; \rm mol[/tex] of [tex]\rm MgSO_4[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}[/tex].

Therefore, the mass of [tex]\rm MgSO_4[/tex] required to make this solution would be approximately [tex]19.1\; \rm g[/tex].


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Answers

Answer:

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Neutron = Neutral charge (=)

Electron = Negative Charge (-)

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Answer:

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Answers

ANSWER:

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What is the mass of CO2 lost at 20 min from the limestone sample?

How much CO2, in moles, was lost at 20 min from the limestone sample?

How much CO3 2- was lost at 20 min from the limestone sample?

Answers

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore  

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO​​​ 3

Hence in sample present CaCO​​​ 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO​​​3

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