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3 years ago

calculate the volume that will be occupied by 350 ml of oxygen measured at 720 mm hg, when the pressure changed to 630 mm hg

Chemistry

1 answer:

likoan [24]3 years ago

8 0

Answer:

406.45mL

Explanation:

The following data were obtained from the question:

V1 = 350mL

P1 = 720mmHg

P2 = 630mmHg

V2 =?

The new volume can be obtain as follows:

P1V1 = P2V2

720 x 350 = 620 x v2

Divide both side by 620

V2 = (720 x 350) /620

V2 = 406.45mL

The new volume of the gas is 406.45mL

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How many moles of CaO will be produced from 23.9 g of Ca?<br>2Ca(s) + O2(g) – 20:0(3)<br>mol

inessss [21]

Answer:

0.6 moles of CaO will produced.

Explanation:

Given data:

Mass of calcium = 23.9 g

Moles of CaO produced = ?

Solution:

Chemical equation:

2Ca + O₂ → 2CaO

Number of moles of calcium:

Number of moles = mass/ molar mass

Number of moles = 23.9 g / 40 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of calcium and CaO.

Ca : CaO

2 : 2

0.6 : 0.6

0.6 moles of CaO will produced.

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3 years ago

The volume (in L) that would be occupied by 5.00 mols of 02 at STP is

crimeas [40]

Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of $O_2$ = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

$PV=nRT$

where,

P = pressure of gas

V = volume of gas

n = the number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K (Given)

By rearranging the above ideal gas law equation, we get

$V=\frac{nRT}{P}$

Now put all the given values in this expression, we get the value of volume.

$V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L$

Therefore, the volume of oxygen at STP is 112.0665 L

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Read 2 more answers

Britney added 0.05 moles of copper(II) nitrate solution to 0.1 moles of sodium hydroxide solution and

Rama09 [41]

The percent yield of copper hydroxide is 84%

<h3>Stoichiometry</h3>

From the question, we are to determine the percent yield of copper hydroxide

First, we will determine the theoretical mass

From the given balanced chemical equation, we have

Cu(NO₃)₂ + 2NaOH -- Cu(OH)₂ + 2NaNO₃

This means,

1 mole of copper(II) nitrate reacts with 2 moles of sodium hydroxide to produce 1 mole of copper hydroxide

Therefore,
0.05 mole of copper(II) nitrate reacts with 0.1 mole of sodium hydroxide to produce 0.05 mole of copper hydroxide

The theoretical number of moles of copper hydroxide that is produced is 0.05 mole

Now, for the theoretical mass

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of copper hydroxide = 97.56 g/mol

Then,

Theoretical mass = 0.05 × 97.56

Theoretical mass of copper of hydroxide produced is = 4.878 g

Now, for the percent yield of copper hydroxide

Percent yield is given by the formula,

$Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100\%$

Then,

$Percent\ yield\ of\ copper\ hydroxide= \frac{4.1}{4.878}\times 100\%$

$Percent\ yield\ of\ copper\ hydroxide= 84\%$

Hence, the percent yield of copper hydroxide is 84%.

Learn more on Stoichiometry here: brainly.com/question/9372758

7 0

3 years ago