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stiks02 [169]
3 years ago
7

Consider 100.0 g samples of two different compounds consisting only of vanadium and oxygen. one compound contains 61.4 g of vana

dium and the other has 76.1 g of vanadium. find the ratio:
Chemistry
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

1. Compound 1

Mass of O = 100.0 - 61.4 = 38.6 g

r_{1} = \dfrac{\text{mass of O}}{\text{mass of V}} = \dfrac{ 38.6}{61.4} = 0.6287

2. Compound 2

Mass of O = 100.0 - 76.1 = 23.9 g

r_{2} = \dfrac{ 23.9}{76.1} = 0.3141

3. Ratio of the ratios

r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.6287}{0.3141} = \dfrac{2.00}{1} \approx 2\\\\\text{The relative amounts of O per gram of V are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}

For example. the compounds might be VO₂ and VO.

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A chemist must dilute of aqueous sodium carbonate solution until the concentration falls to . He'll do this by adding distilled
Mademuasel [1]

The question is incomplete, complete question is :

A chemist must dilute 73.9 mL of 400 mM aqueous sodium carbonate  solution until the concentration falls to 125 mM . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.

Answer:

The final volume of the solution will be 0.236 L.

Explanation:

Concentration of sodium carbonate solution before dilution =M_1= 400 mM

Volume of sodium carbonate solution before dilution = V_1=73.9 mL

Concentration of sodium carbonate solution after dilution =M_2= 125 mM

Volume of sodium carbonate solution after dilution = V_2=?

Dilution equation is given by:

M_1V_1=M_2V_2

V_2=\frac{M_1V_1}{M_2}

V_2=\frac{400 mM\times 73.9 mL}{125 mM}= 236.48 mL\approx 236 mL

1 mL = 0.001 L

236 mL = 0.236 L

The final volume of the solution will be 0.236 L.

3 0
3 years ago
Radium-223 has a half-life of 11.0 days. How much of a 60.0 mg sample of Ra-223 is still active after 44.0 days?
Airida [17]

Answer:

The answer to your question is 3.75 mg

Explanation:

Data

Radium-223

half-life = 11 days

mass = 60 mg

total time = 44 days

Process

1.- Draw a chart to start the calculation

                Mass           Number of days      Half-lives

                60 mg                    0                           0

                30 mg                    11                            1

                 15 mg                    11                            2

                7.5 mg                    11                           3

               3.75 mg                   11                           4

        Total number of days   44

       Mass active after 44 days = 3.75 mg                                      

           

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3 years ago
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OleMash [197]

Answer:

precipitation is when water flows into the ground

Explanation:

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