A gas at a constant volume has an initial temperature of 300.0 K and an initial pressure of 10.0 atm. What pressure does the gas

If matter is uniform throughout, cannot be separated into other substances by physical processes, but can be decomposed into oth

olga nikolaevna [1]

D. because matter is uniform through and cannot be separated into other substances by physical means.

6 0

3 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

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$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

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To find molality, we have to divide no. of moles of solute by weight of solution

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While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

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Now, we will find molality

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$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

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$\textsf{ \large{ \underline{Now substituting the required values}}}$

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$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

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Henceforth, the change in boiling point is 0.00735°C.

7 0

1 year ago

9. A student wished to prepare ethylene gas by dehydration of ethanol at 140oC using sulfuric acid as the dehydrating agent. A l

Arte-miy333 [17]

We have that the the liquid is

C_2H_5OH (ethanol
And at a condition of H_2SO4 as catalyst and temp 170

From the question we are told

A student wished to prepare ethylene gas by dehydration of ethanol at 140oC using sulfuric acid as the dehydrating agent.
A low-boiling liquid was obtained instead of ethylene.
What was the liquid, and how might the reaction conditions be changed to give ethylene

<h3>Ethylene formation</h3>

Generally the equation is

2C_2H_5OH------CH3CH_2O-CH_2CH_3+H_20

Therefore

with ethanol at 140oC

The product is diethyl ethen

The reaction at 170 ethylene will give

C_2H_5OH-------CH_2=CH_2+H_2O( at a condition of H_2SO4 as catalyst and temp 170)

Therefore

The the liquid is

C_2H_5OH (ethanol

For more information on Ethylene visit

brainly.com/question/20117360

8 0

2 years ago

The conversion of 37 ˚C to K

77julia77 [94]

Answer:

310.15

Explanation:

c + 273.15 = k

37+ 273.15 = 310.15

8 0

3 years ago

Read 2 more answers

Si se adiciona etilenglicol (compuesto no volátil) a agua destilada:

poizon [28]

Answer:

Hey hi

Explanation:

Can you pls tell me which language is this pls I really want to help you... Sorry

6 0

3 years ago