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3 years ago

What amount of moles does 60.5 L of carbon dioxide occupy at 2.15 atm and 25 degrees C?

Chemistry

1 answer:

xeze [42]3 years ago

8 0

Answer: 5.3 moles

Explanation:

Hi, to answer this question we have to apply the next formula:

n = PV/RT

Where:

n = number of moles

P= Pressure = 2.15 atm

V= volume = 60.5 L

R = 0.0821 atm(L)/mol(K)

T = temperature = 25 °+ 273.15 = 298.15 Kelvin

Replacing with the values given:

n = 2.15 (60.5) / 0.0821 (298.15) = 5.3 moles.

Feel free to ask for more if needed or if you did not understand something.

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PLEASE HELP!?Consumers must eat other organisms for energy. Which organisms are consumers in this food chain?

Vitek1552 [10]

Answer:

Consumers must consume other organisms to get the food that they need and are known as Heterotrophs as they cannot make their own glucose. These consumers eat producers (plants). Herbivores are considered as first order consumers. These consumers eat consumers and producers (animals and plants).

4 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

⠀

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

⠀

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.

⠀

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

⠀

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)

⠀

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

⠀

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

⠀

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

⠀

While first we need to no. of moles

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

⠀

Now, we will find molality

⠀

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