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3 years ago

A gas occupies 18.7 L at standard temperature. What is its volume at 200.5 °C

Chemistry

1 answer:

nexus9112 [7]3 years ago

7 0

Answer:

32.4 L

Explanation:

Note 1:

The Standard Temperature and Pressure (STP) is defined by IUPAC as air at 0 gradius Celcius and 1 bar.

Note 2:

The Ideal Gas Law is used for solving this problem.

......

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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Which statement is true about the exothermic reaction?

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Explanation:

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3 years ago

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3 years ago

Which solution has the higher boiling point?

dedylja [7]

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>

The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
The definition of molarity, on the other hand, is based on a certain volume of solution.
Mol/kg is a typical molality measurement unit in chemistry.

<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

Learn more about boiling point here:

brainly.com/question/24168079

#SPJ4

7 0

2 years ago

The half-life of gold-198 is 2.7 days. After

Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= $\frac{x}{4}$

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{2.7days}=0.257days^{-1}$

b) for completion of one fourth of reaction

$t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}$

$t=\frac{2.303}{0.257}\log{4}$

$t=8.1days$

Thus after 8.1 days , one fourth of original amount will remain.

8 0

3 years ago