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3 years ago

How many grams of P4O10 would be produced when 0.700 mole of phosphorus is burned?

1 answer:

8 0

We can solve the equation and show the solution below:

Oxygen atomic number is 16.
Phosphorus atomic number is 32.

We have the molecular weight:
Molecular weight = (31*4) + (16*10)
Molecular weight = 284 grams/mol

Solving for the grams:

0.4 mole (for P4) * (1 mol P4O10/1 mol P4) * (284 grams P4O10/1 mole P4O10)
Total grams = 113.6

The answer is 113.6 grams.

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At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L . What volume wo

Nostrana [21]

Taking into account the Charles's law, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

<h3>Charles's Law</h3>

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, at a constant pressure.

Volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law is a law that says that the quotient that exists between the volume and the temperature will always have the same value:

V÷ T= k

Considering an initial state 1 and a final state 2, it is satisfied:

V1÷ T1= V2÷ T2

<h3>Volume at 65°C</h3>

In this case, you know:

V1= 22.4 L
T1= 0 C= 273 K
V2= ?
T2= 65 C= 338 K

Replacing in Charles's law:

22.4 L÷ 273 K= V2÷ 338 K

Solving:

(22.4 L÷ 273 K) ×338 K= V2

Finally, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

Learn more about Charles's law:

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3 0

1 year ago

Someone help me am i right or wrong

steposvetlana [31]

Answer:

You right!

Explanation:

3 0

3 years ago

Read 2 more answers

Calculate the molarity of 139 grams of sucrose, C12H22O11, in 2.60 liters of solution.

Allisa [31]

Answer:

to calculate the molarity of the said sucrose,

firstly calculate the moles

which is = Molecular weight of C12H22O11 = 342g/mol

then

moles = 139/342

= 0.41 moles

to calculate Molarity now

Molarity= moles of the solute/volume of solution in liter

=0.41/2.60

=0.158M

Explanation:

4 0

3 years ago

Five calcite, CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 12.2±0.1 g . What is the average mass and abs

Westkost [7]

Answer:

<em>The average mass of calcium in each sample is: </em><u>0.978 g</u>

<em />

<em>The absolute uncertainty is: </em><u>0.008 g</u>

Explanation:

The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.

When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.

The relative uncertainty is the absolute uncertainty divided by the quantity:

Relative uncertainty = 0.1g / 12.2 g = 0.008

The average mass of calcium is calculated using proportions, along with the molar masses:

Molar mass of calcium: 40.078 g/ mol (from a periodic table)

Molar mass of calcite: 100.085 g/mol (given)

Proportion:

40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite

x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium

So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:

Average mass = total mass of five samples / number of samples

Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>

So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).

The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:

Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g

The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.

7 0

3 years ago

really need this answer

scoundrel [369]

Answer:

B 100,000%

Explanation:

3 0

2 years ago