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1 year ago

Calculate the mass, in grams, of 2.74 l of co gas measured at 33°c and 945 mmhg.

1 answer:

4 0

<span>T = 33 + 273 =306 K moles CO = pV/RT = 1.24 x 2.74 / 0.08206 x 306=0.135 mass CO = 0.135 mol x 28.01 g/mol=3.78 g</span>

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At this point Ron is slightly confused, this isn’t surprising. However, Hermione is doing rather well with them. This also isn’t

Zigmanuir [339]

Answer:

$\boxed{\text{0.780 atm}}$

Explanation:

Hermione is pretty smart. She realizes that, according to Dalton's Law of Partial Pressures, each gas exerts its pressure independently of the others, as if the others weren't even there.

She shows Ron how to use the Ideal Gas Law to solve the problem.

pV = nRT

She collects the data:

V = 1.00 L; n = 0.0319 mol; T = 25.0 °C

She reminds him to convert the temperature to kelvins

T = (25.0 +273.15) K = 298.15 K

Then she shows him how to do the calculation.

$p \times \text{1.00 L} = \text{0.0319 mol} \times \text{L}\cdot\text{atm}\cdot\text{0.082 06 K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\\\1.00p = \text{0.7805 atm}\\\\p = \textbf{0.780 atm}\\\\\text{The partial pressure of the nitrogen is } \boxed{\textbf{0.780 atm}}$

Isn't she smart?

4 0

2 years ago

Which statement is TRUE regarding the macroscopic and

damaskus [11]

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

7 0

1 year ago

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

ladessa [460]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

1 year ago

Which type of reaction occurs when a high energy particle collides with the nucleus of an atom, converting that atom to an atom

Finger [1]

The term used is transmutation

3 0

2 years ago

Read 2 more answers

Competition is likely to occur between which two organisms?

MatroZZZ [7]

<h3>Full question found from another source:</h3>

5. Competition is likely to occur between which two organisms?

A Owls and bacteria

B Deer and butterflies

C Grass and strawberry plants

D Goldfish and rabbits

Answer:

Explanation:

Grass and strawberry plants compete for the same resources such as nutrients, water, habitat, and sunlight.

Bacteria and owls rely on completely different resources and mechanisms to survive - bacteria are single cell organisms and owl are complex birds

Deer and butterflies are also are not in competition with eachother - they don't occupy the same habitats, eat the same food or have the same prey.

Goldfish and rabbits will never encounter eachother, given that one lives in water and one lives on land. They do not compete for the same food source or habitat.

7 0

1 year ago