What happens to carbon compounds in air when heated sufficiently?

47.9 ml hrdrogen is collected at 26° Celsius and 718 torr. Find the volume occupied at STP

Temka [501]

Answer:

41.45 mL

Explanation:

Applying the general gas equation,

PV/T = P'V'/T'............... Equation 1

Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.

make V' the subject of the equation

V' = PVT'/TP'................ Equation 2

Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²

Substitute these values into equation 2

V' = ( 95725.196×0.0479×273)/(299×101000)

V' = 0.04145 dm³

V' = 41.45 mL

4 0

3 years ago

Hurry and Help for Brainlist

zalisa [80]

Ummm i think it’s solid

3 0

2 years ago

Read 2 more answers

Given that a mass of a ball is (36.0 g), and the volume is (8.0 mL), what is the ball's density in g/mL?

max2010maxim [7]

Answer:

4.5g/mL

Explanation:

Given parameters:

Mass of ball = 36g

Volume of the ball = 8mL

Unknown:

Density of the ball = ?

Solution:

Density is the mass per unit volume of a substance.

Density = $\frac{mass}{volume}$

So;

Density = $\frac{36}{8}$ = 4.5g/mL

5 0

2 years ago

so they are asking me to find the empirical formula for a compound that is 7.70% carbon and 92.3% chlorine. Can you show me step

Nikolay [14]

Answer:

Empirical formula is CCl₄

Explanation:

Given data:

Percentage of carbon = 7.70%

Percentage of chlorine = 92.3%

Empirical formula = ?

Solution:

Number of gram atoms of Cl = 92.3 / 35.5 = 2.6

Number of gram atoms of C = 7.70 / 12 = 0.64

Atomic ratio:

C : Cl

0.64/0.64 : 2.6/0.64

1 : 4

C : Cl = 1 : 4

Empirical formula is CCl₄.

5 0

3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

6 0

2 years ago