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EleoNora [17]
3 years ago
15

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

tion is completed; calculate the concentration of nitrate ions in solution. Assume that the total volume of the solution is 3.0 x 10^2 mL
Chemistry
1 answer:
Murljashka [212]3 years ago
6 0

Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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SashulF [63]
<h3>Answer:</h3>

0.144 moles

<h3>Explanation:</h3>
  • The relationship between mass of a compound, number of moles and molar mass of the compound is given by;
  • Number of moles = Mass ÷ Molar mass
  • Molar mass is equivalent to the relative formula mass of the compound that is calculated the atomic masses of the elements making the compound.

In this case;

Our compound, KClO3 will have a molar mass of;

= 39 + 35.5 + 4(16)

= 138.5 g/mol

Mass of KClO3 is 20 g

Therefore;

Number of moles = 20 g ÷ 138.5 g/mol

                            = 0.144 moles

Thus, the number of moles in 20 g of KClO3 is 0.144 moles

4 0
3 years ago
True or false: a physical property is a characteristic of matter that you cannot observe or measure without changing the identit
Ganezh [65]
That is false, hope this helps.
4 0
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Romashka-Z-Leto [24]

Answer:

Nucleotides

Explanation:

Nucleotides are the organic molecules which serve as monomer units for the formation of nucleic acid polymers which are the deoxyribonucleic acid and the ribonucleic acid (RNA) and both are the essential biomolecules within the life on the Earth.

Nucleotides are building blocks of the nucleic acids. They are the molecules which are composed of three sub units which are:  

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8 0
3 years ago
Methane burns in the presence of oxygen to form carbon dioxide and water.
wlad13 [49]

Answer:

= 9.28 g CO₂

Explanation:

First write a balanced equation:

CH₄ + 2O₂ -> 2H₂O + CO₂

Convert the information to moles

7.50g CH₄ = 0.46875 mol CH₄

13.5g O₂ = 0.421875 mol O₂

Theoretical molar ratio CH₄:O₂ -> 1:2

Actual ratio is  0.46875 : 0.421875 ≈ 1:1

If all CH₄ is used up, there would need to be more O₂

So O₂ is the limiting reactant and we use this in our equation

Use molar ratio to find moles of CO₂

0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂

Then convert to grams

0.2109375 mol CO₂ = 9.28114 g CO₂

round to 3 sig figs

= 9.28 g CO₂

5 0
3 years ago
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I would assume so.

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Both would obtain the same proportions, so I don't see why putting a half cup of sugar would make things any different.

Hope this is the answer you are looking for.
8 0
3 years ago
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