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1 year ago

How many mL of a 4.50M NaBr solution are needed to make 250 mL of a 0.75M solution of NaBr?

1 answer:

8 0

M1V1=M2V2

(4.50)(x)=(0.75)(250)
(4.50)(x)=187.5
x=41.67 mL
You might want to double check the formula used, but as a chem student I’m confident this is right

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Describe an experience you've had making or building something where the amount of each ingredient or building block came in fix

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If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?

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A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

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Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

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2 years ago

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1 year ago

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

Sergio039 [100]

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n = m/M
63.57 g C -> 63.57 g C/12.01 g/mol = 5.29 moles C
6 g H -> 6 g C/1.008 g/mol = 5.95 moles H
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So the minimum formula has this rate:

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Let’s multiply all numbers by 2:

C8H9N1O2

<span>Now they are all whole numbers!
</span>

So the minimum formula is C8H9NO2

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