All categories

2 years ago

What is the density of 2.5 g of gaseous sulfur held at 130. kPa and 10.0 degrees Celsius?

2 answers:

7 0

Assuming that the gaseous sulfur will behave as an ideal gas,
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(101.3 x 22.4)/(273) = (130 x V₂)/(283)
V₂ = 18.1 L
density = mass/volume
density = 2.5/18.1
= 0.14 g/L

worty [1.4K]2 years ago

4 0

Use ideal gas equation: pV = nRT

Now pass n to mass: n = mass / MM .... [MM is the molar mass]

pV = [mass/MM]*RT =>mass/V = [p*MM] / RT and mass / V = density

p= 130 kPa = 130,000 Pa = 130,00 joule / m^3
T = 10.0 ° + 273.15 = 283.15 k
MM of sulfur (S) = 32 g/mol = 32000 kg/mol

density = 130,000 Pa * 32000kg/mol / [8.31 joule / mol*k * 283.15 k] = 1.77*10^6 kg/m^3 = 1.77 g/L ≈ 1.8 g/L

Then, I do not get any of the option choices.

Is it possbile that the pressure is 13.0 kPa instead 130. kPa? If so the answer would be 18 g/L

Note that the mass is not used. You do not need it unless you are asked for the volume, which is not the case.

You might be interested in

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Kisachek [45]

Answer: The empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

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 $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 39.01 g of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 10.65 g of water, $\frac{2}{18}\times 10.65=1.18g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is $C_9H_{12}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Hence, the empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

6 0

2 years ago

Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a

Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

3 0

2 years ago

Calculate the number of chloride ions in 6.8 g of zinc chloride

Bumek [7]

Answer:

The correct answer is 0.300 * 10^23 ions.

Explanation:

Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.

The moles of zinc chloride (ZnCl2) is,

= mass of zinc + 2 mass of chlorine

= 65.38 + 2 (35.45)

=65.38 + 70.90

= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)

Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine

Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8

= 3.537 g of chlorine

70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537

= 0.300 * 10^23 ions of chlorine.

3 0

2 years ago

To 100.0 g water at 25.00 ºc in a well-insulated container is added a block of aluminum initially at 100.0 ºc. the temperature o

11111nata11111 [884]

When the amount of heat gained = the amount of heat loss

so, M*C*ΔTloses = M*C* ΔT gained

when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.

∴( M * C * ΔT )W = (M*C*ΔT) Al

when Mw is the mass of water = 100 g

and C the specific heat capacity of water = 4.18

and ΔT the change in temperature for water= 28-25 = 3 ° C

and ΔT the change in temperature for Al = 100-28= 72°C

and M Al is the mass of Al block

C is the specific heat capacity of the block = 0.9

so by substitution:

100 g * 4.18*3 = M Al * 0.9*72

∴ the mass of Al block is = 100 g *4.18 / 0.9*72

= 19.35 g

4 0

2 years ago

A balance measures mass to 0.001 g. If you determine the mass of an object that weighs about 30 g, would you record the mass as

solong [7]

Answer:

The mass is recorded as 32.075 g

Explanation:

"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the whole figure.

8 0

2 years ago

Read 2 more answers