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ryzh [129]
3 years ago
5

A student dissolved 3.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are in the final solution?
Chemistry
1 answer:
viktelen [127]3 years ago
5 0

Answer:

0.081g

Explanation:

first step:calculate the molar conc of the stock solution

molar conc=mass/molar mass ×1000/vol(ml)

molar mass of Co(N03)2=182.943

molar conc=3g/182.943 ×1000/100ml

=0.163M

To calculate the conc of the resulting solution we apply the dilution principle

n=cv

C1V1=C2V2

C1=conc of stock solution=0.163M

V1=4mL

V2=275ml(final volume)

by making 'C2" the subject C2=0.163×4/275

=0.0024M

by applying the formular for calculating molar conc of a solution,we can calculate the mass of CO(NO3)2 in the reulting solution

molar conc=mass/molar mass ×1000(L)/vol(mL)

mass=0.0024×182.943×275/1000

=0.1193g

Mass of CO(NO3)2  in the resulting solution=0.1199g

since CO+(NO3)2=CO(NO3)2

Percentage by mass of (NO3)2 in CO(NO3)2=(NO3)2/CO(NO3)2

=124/182.943×100

=67.8%

it means 67.8% of 0.1193g is the mass of (NO3)2=0.678×0.1199

=0.081g

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7nadin3 [17]

Answer:

<h3>The answer is 7.42 </h3>

Explanation:

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From the question we have

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We have the final answer as

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4 0
2 years ago
What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?
My name is Ann [436]

Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

  • 175mg/times 1g/1,000mg=0.175g

  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

a) <u>For CO₂</u>

i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
  • mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

  • mass = number of moles × molar mass
  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
4 0
3 years ago
The density of pure solid copper is 8.94 g/mL. What volume does 34 g of copper occupy?
patriot [66]

hey there ! :

Density = 8.94 g/mL

mass = 34 g

Volume = ??

Therefore:

D = m / V

8.94 = 34 / V

V = 34 / 8.94

V = 3.803 mL

Hope this helps!

4 0
3 years ago
A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts
Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

                    1 quarts = 0.95 liter

                   7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

        Given:

            1 liter of solution contains 61.2g of barium nitrate:

          7.125 liter will contain  7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

  In solution we have  Ba²⁺ + NO₃⁻

                Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

 

Number of moles of Ba(NO₃)₂ = \frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = \frac{436.05}{271} = 1.609mole

      1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

    therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions  x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0
3 years ago
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