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DiKsa [7]
3 years ago
10

Convert 6.71 moles NH3 to number of molecules.

Chemistry
1 answer:
svetlana [45]3 years ago
6 0

Answer: There are 40.41 \times 10^{23} molecules present in 6.71 moles of NH_{3}.

Explanation:

Given: Moles of NH_{3} = 6.71 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{22} molecules.

Therefore, molecules present in 6.71 moles are calculated as follows.

Molecules = moles \times 6.022 \times 10^{23}\\= 6.71 \times 6.022 \times 10^{23}\\= 40.41 \times 10^{23}

Thus, we can conclude that there are 40.41 \times 10^{23} molecules present in 6.71 moles of NH_{3}.

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Below are the steps of the engineering process (not necessarily in order): 1) Know the background 2) Do the work 3) Make a plan
LuckyWell [14K]

Answer:

4 - 1 - 3 - 2 - 6 - 5

Explanation:

During an engineering process, first, we need to identify the problem, or the need because the process only will occur because of some need. Then, it's necessary to know as much as possible about the problem and the things that already exist or already were tested to solve it. Knowing the background will make the work easy.

After that, it's necessary to plan the things we'll do, knowing the costs, the time needed for activities, how many people will be necessary for each step, etc. It's really important to make a plan. Then, do the work, following the plan. Thus, the process must be tested. During the test of the results, some problems must be found, so it's time to evaluate and redesign the process, to solve these problems found.

3 0
3 years ago
Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,
sammy [17]

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

  • Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

  • Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

  • ΔT is the boiling-point elevation
  • Kb is the ebullioscopic constant.
  • b is the molality.
  • i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

7 0
2 years ago
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

  • Partial pressure of methane (pCH₄): 431 mmHg
  • Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0
2 years ago
HELP ASAP 20 POINTS AAAA
pochemuha

Answer:

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So then Molly mixed the baking soda and vinegar together and checked the TEMPERATURE. We all OBSERVED as the thermometer’s TEMPERATURE went down. “ your THEORY/ HYPOTHESIS was correct!” Exclaimed Molly.

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Explanation:

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3 0
2 years ago
How many molecules of CaCi2 are equivalent to 75.9 grams
horsena [70]
<em>M CaCl₂: 40+(35,5×2) = 111 g/mol</em>


6,02·10²³ molecules ---------- 111g
X molecules --------------------- 75,9g
X = (75,9×<span>6,02·10²³)/111
X = <u>4,116</u></span><span><u>·10²³</u> molecules of CaCl</span>₂

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3 years ago
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