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ASHA 777 [7]
3 years ago
8

Calculate the volume of 7 mol of hydrogen at stp

Chemistry
1 answer:
dimulka [17.4K]3 years ago
5 0
Volume of hydrogen at STP = moles x 22.4 = 7x22.4 = 156.8L
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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol
katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of 0.80^oC and a freezing point depression constant K_f=7.82^oC.kg/mol . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is -17.6^oC

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

where,

Freezing point of pure solvent = 0.80^oC

Freezing point of solution = ?^oC

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

K_f = freezing point depression constant = 7.82^oC/m

m_{solute} = Given mass of solute (iron (III) chloride) = 81.1 g

M_{solute} = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

w_{solvent} = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC

Hence, the freezing point of the solution is -17.6^oC

6 0
2 years ago
Please help
professor190 [17]

Answer:

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6 0
3 years ago
Which is the strongest acid? Ch3ch2ch2chfch2co2h ch3ch2ch2cbr2ch2co2h ch3ch2ch2cf2ch2co2h ch3chbrch2ch2ch2co2h ch2clch2ch2ch2ch2
liq [111]

The strength of an acid increases if the stability of conjugate base increases

The stability of a conjugate base increases with the presence of electron with drawing group (electronegative group)

Thus more the electronegativity of an atom attached to a carboxylic acid higher the strength of acid

In these examples CH3CH2CH2CF2CH2COOH contains to electronegative flourine atoms which stabilizes the conjugate base hence this will be the strongest acid among the given acids

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How do purple stem plants use there energy
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Purple stem plants can be formed by genetics, they use their energy slightly different from other plants, they use less energy, but that may also be because of bad nutrition, and because they may be hungry for nutrients.
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