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HACTEHA [7]
2 years ago
15

6.How many moles of gas would be in contained in a 11.2 L container that is at a

Chemistry
1 answer:
qaws [65]2 years ago
4 0

0.34 moles of gas would be contained in a 11.2 L container that is at a pressure of 0.75 atm and 300 K.

<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>

The number of moles of a substance can be calculated using the following expression:

PV = nRT

Where;

  • p = pressure (atm)
  • v = volume (L)
  • n = number of moles
  • R = gas law constant
  • T = temperature

0.75 × 11.2 = n × 0.0821 × 300

8.4 = 24.63n

n = 8.4 ÷ 24.63

n = 0.34 moles

Therefore, 0.34 moles of gas would be contained in a 11.2 L container that is at a pressure of 0.75 atm and 300 K.

Learn more about number of moles at: brainly.com/question/1190311

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According to our textbook, write the formula for the anhydrous compound that was part of the mixture called natron that was used
bearhunter [10]

Answer:

The formula for the anhydrous compound that was part of the mixture called natron that was used by the Egyptians is Na2(CO3)10(H2O).

They use this compound for medicine, cookery, agriculture, in glass-making and to dehydrate egyptian mummies.

Compound of sodium carbonate and sodium bicarbonate was the name of the resulting hydrate that formed.

8 0
3 years ago
a. What is the definition of a mole? b. What is the abbreviation for mole? SARM c. How many particles are in one mole? d. What n
ycow [4]

Mole is base unit of substance. The number of particles in one mole is equal to 6.022 × 10²³. The number of particle in a mole is called as Avogadro's number.

The mole of substance is the amount of substance that contain elementary particles as the number of atoms. the symbol of mole is mol. it is the SI unit of amount of substance. The number particles are ion one mole of substances is 6.022 × 10²³. The number of particles in one mole of substance is called as Avogadro's number.

Thus, Mole is base unit of substance. The number of particles in one mole is equal to 6.022 × 10²³. The number of particle in a mole is called as Avogadro's number.

To learn more about moles here

brainly.com/question/26416088

#SPJ1

7 0
1 year ago
The structural level of a protein least affected by a disruption in hydrogen bonding is the
sergeinik [125]
I think the correct answer from the choices listed above is option A. The structural level of a protein least affected by a disruption in hydrogen bonding is the primary level. The other levels are very much affected by hydrogen bonding. Hope this answers the question.
7 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
Hey lol hey lol hey lol hey lol
Elanso [62]

Answer:

Hi!!!

Explanation:

Hello hope you are having a great day:)

5 0
3 years ago
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