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zhannawk [14.2K]
2 years ago
9

Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.

Chemistry
1 answer:
Travka [436]2 years ago
5 0

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means  1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × \frac{254 g}{1 mole}= 508 g of Phenobarbital sodium.

1 L ×\frac{1000 mL}{ 1 L} = 1000 mL of solution

Thus, % w/v is:

\frac{508 g}{1000 mL} × 100 = 50,8 % w/v

And radio strengths:

\frac{1000 mL}{508 g}  = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

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A 7.591-9 gaseous mixture contains methane (CH4) and butane
mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y <em>(1)</em>

<em>Where X is the mass of methane and Y the mass of butane</em>

<em />

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

<em>2 moles of oxygen react per mole of methane</em>

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

<em>13/2 moles of oxygen react per mole of methane</em>

<em />

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y <em>(2)</em>

<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

<h3>65.71%</h3>

7 0
2 years ago
A blacksmith making a tool heats 525 grams of steel to 1230°C. After hammering the steel, she places it into a bucket of water t
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A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
Grace [21]
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
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