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butalik [34]
3 years ago
14

This is a chemistry question that i don't know how to do. Need some help

Chemistry
1 answer:
koban [17]3 years ago
5 0
C₀=8.10M
c₁=5.28M
v₀=1.58 L
v₁-?

n=c₀v₀=c₁v₁

v₁=c₀v₀/c₁

v₁=8.10*1.58/5.28=2.42 L 
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The "normal" lead content in human blood is about 0.40 part per million (i.e., 0.40 g of lead per million grams of blood). A val
Reil [10]

Answer:

in 6.0 E3 g of blood at a concentration of 0.62 ppm there are  3.5g of lead content

Explanation:

  • blood density: 1.06 g/L
  • ppm ≡ mg Pb / L blood

⇒ V blood = 6.0E3gblood * L / 1.06gblood = 5660.37 L blood

<u> mass of lead:</u>

⇒ 5660.37 L blood * 0.62 mg Pb / L blood * 0.001 g / mg

⇒ mass = 3.509 g of lead

7 0
4 years ago
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A 12.41 g sample of nabr contains 22.34% na by mass. considering the law of constant composition (definite proportions) how many
Helga [31]
I'm assuming you mean how many grams of sodium does 8.99 grams of sodium bromide have, because Bromine would not have any sodium in it.

Then, just direct calculate 22.34
(22.34 \div 100) \times 8.99 = 2.008366
6 0
3 years ago
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Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and
Gre4nikov [31]

Explanation:

Relation between entropy change and specific heat is as follows.

            \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

The given data is as follows.

     mass = 500 g,         C_{p} = 24.4 J/mol K

     T_{h} = 500 K,          T_{c} = 250 K               

   Mass number of copper = 63.54 g /mol

Number of moles = \frac{mass}{/text{\molar mass}}

                                 = \frac{500}{63.54}

                                 = 7.86 moles

Now, equating the entropy change for both the substances as follows.

     7.86 \times 24.4 \times [T_{f} - 250] = 7.86 \times 24.4 \times [500 -T_{f}]

       T_{f} - 250 = 500 - T_{f}

          2T_{f} = 750

So,       T_{f} = 375^{o}C

  • For the metal block A,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

              = 24.4 log [\frac{375}{500}]

              = -3.04 J/ K mol

  • For the block B,  change in entropy is as follows.

         \Delta S = C_{p} log (\frac{T_{2}}{T_{1}})

                  = 24.4 log [\frac{375}{250}]

                  = 4.296  J/Kmol

And, total entropy change will be as follows.

                       = 4.296 + (-3.04)

                      = 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol  and change in entropy of block B is 4.296  J/Kmol.

8 0
3 years ago
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vladimir1956 [14]

Answer:

A. Stable

Explanation:

unless ur talking about the star in your PFP, then it's unstable xd

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tino4ka555 [31]
(a) sodium hydroxide + ammonium sulfate --> sodium sulfate + ammonium hydroxide

sodium sulfate precipitates in aqueous solutions

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</span>(c) strontium bromide + silver nitrate ---> strontium nitrate + silver bromide

both  do not precipitate in aqueous solution
3 0
3 years ago
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