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Rzqust [24]
3 years ago
6

A stock solution containing Mn2 ions was prepared by dissolving 1.269 g pure manganese metal in nitric acid and diluting to a fi

nal volume of 2.000 L. The following solutions were then prepared by dilution: For solution A, 40.00 mL of stock solution was diluted to 1500.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, 20.00 mL of solution B was diluted to 550.0 mL. a Calculate the concentration of the stock solution.
Chemistry
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

B.

Explanation:

Hope this helpssss

<33

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stepan [7]

Answer:

la parte de la ciencia que se ocupa de las sustancias que componen la materia, la investigación / estudio de sus propiedades y reacciones, y la implementación de tales reacciones para formar nuevos productos / sustancias.

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4 0
3 years ago
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The carbon−14 decay rate of a sample obtained from a young tree is 0.266 disintegration per second per gram of the sample. Anoth
Novosadov [1.4K]

Answer:

  • <u><em>3.30 × 10³ years</em></u>

Explanation:

1) <u>Carbon-14 disintegration rate of a young (alive) tree</u>: 0.266 per second per gram.

2)<u> Carbon-14 disintegration rate of a wood sample prepeared from an object recovered at an archaelogical excavation (dead matter):</u> 0.178 per second per gram.

3) <u>Ratio of decay</u>: A/A₀ = 0.178 / 0.266 = 0.669

4) <u>Half-life equation</u>: A/A₀ = (1/2)ⁿ, where n is the number of half-lives since the object died.

5) <u>Substitute and solve for n</u>:

  • 0.669 = (1/2)ⁿ

nlog(1/2)=log(0.669)\\ \\n=\frac{log(0.669)}{log(1/2)}\\ \\n=0.578

That means that 0.578 half-life has elapsed since the wood with which the object was created was dead matter.

6) <u>Convert number of half-lives to years</u>:

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5 0
3 years ago
A solution containing a mixture of 0.0333 M0.0333 M potassium chromate ( K2CrO4K2CrO4 ) and 0.0532 M0.0532 M sodium oxalate ( Na
jasenka [17]

Answer:

Explanation:

K₂CrO₄ + ( COONa )₂ + 2BaCl₂ = Ba CrO₄ + ( COO ) ₂ Ba + 2 KCl + 2 NaCl

.033 M             .053 M    

Ksp of   Ba CrO₄  is      2.10×10⁻¹⁰

Ksp of ( COO ) ₂ Ba  is   1.30×10⁻⁶

A ) Ksp of   Ba CrO₄ is less so it will precipitate out first .

B) Ksp = 2.10×10⁻¹⁰

Ba CrO₄ = Ba⁺²  +  CrO₄⁻²

                   C            .033

C x   .033  = 2.10×10⁻¹⁰

C =  63.63 x 10⁻¹⁰ M

Ba⁺² must be present in concentration = 63.63 x 10⁻¹⁰ M

C)

90% of precipitation of barium oxalate

concentration of oxalate to precipitate out = .9 x .0532 = .04788

( COO ) ₂ Ba  =  (COO)₂⁻²     +       Ba⁺²

                           .04788 M            C  

C x  .04788  =  1.30×10⁻⁶

C = 27.15  x 10⁻⁶ M .

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3 years ago
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Oxygen, should be correct
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The answer would be choice B because the energy decreased by 20 J
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