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Nonamiya [84]
3 years ago
8

How many moles of O are there in 41.8 grams of O?

Chemistry
1 answer:
il63 [147K]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 2.61 \ mol \ O}}

Explanation:

To convert from grams to moles, the molar mass is used. This number tells us the grams per mole of a substance. It can be found on the Periodic Table. Look for oxygen.

  • Oxygen (O): 15.999 g/mol

Use this number as a ratio.

\frac {15.999 \ g \ O }{ 1 \ mol \ O }

Multiply by the given number of moles.

41.8 \ g \ O *\frac {15.999 \ g \ O }{ 1 \ mol \ O }

Flip the fraction so the grams of oxygen cancel.

41.8 \ g \ O *\frac { 1 \ mol \ O }{ 15.999 \ g \ O}

41.8  *\frac { 1 \ mol \ O }{ 15.999 }

\frac { 41.8 \ mol \ O }{ 15.999 }= 2.61266329146 \ mol \ O

The original measurement of grams has 3 significant figures (4, 1, and 8). Our answer must have the same. For the number we calculated, that is the hundredth place. The 2 in the thousandth place tells us to leave the 1.

2.61 \ mol \ O

41.8 grams of O has 2.61 moles of oxygen.

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as alkali is a salt used as base

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The yeast has the ability to undergo the process of anaerobic respiration and its end product are alcohol, carbon dioxide and 2 moles of ATP. The yeast is grown on maltose medium but unable to produce alcohol because of the presence of oxygen in the medium. The oxygen might acts as poison or inhibit the process of anaerobic respiration.

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If you need 0.0592 moles of nitrogen, how many grams of nitrogen do you need to mass(weigh) on the
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3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0
3 years ago
A 5.00 L sample of helium expands to 12.0 L at which point the
mina [271]

Answer:

1.73 atm

Explanation:

Given data:

Initial volume of helium = 5.00 L

Final volume of helium = 12.0 L

Final pressure = 0.720 atm

Initial pressure = ?

Solution:

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

P₁ × 5.00 L = 0.720 atm × 12.0 L

P₁ = 8.64 atm. L/5 L

P₁ = 1.73 atm

7 0
3 years ago
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