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2 years ago

2.8 dm3 of gas X at STP has equal mass with 11.2 dm3 of C2H6 at STP. What is

Chemistry

1 answer:

ss7ja [257]2 years ago

8 0

Answer: o.125 moles of gas X with a molar mass of 240

Explanation:

2.8 dm3 of gas X at STP has equal mass with 11.2 dm3 of C2H6 at STP. What is the number of mole if gas X has 30 grams?

the molar volume of any gas at STP is the volume of 1 mole at STP

and is 22.4 L

the C2H6 has a volume of 11.2L at STP so it has 0.5 moles

C2H6 has a molar mass of (2 X 12) +(6 X 1) = 24 + 6 = 30

0,5 moles weighs 15 gm at STP with

so 15 gm at STP with 2.8 L for gas X

PV=nRT

P=1 atm

T =273.2k

V =2.8

R =0.082

n = PV/RT = (1 X 2.8)/(0.082 X 273.2) =0.125 moles of gasX

the weight of gas X is 30 gms

so 30/.125 =240 g/mole = molar mass of gas X

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Some COCl2 is placed in a sealed flask and heated to 756 K. When equilibrium is reached, the flask is found to contain COCl2 (7.

o-na [289]

Answer:

$9.044\times 10^{-3}$ is the value of the equilibrium constant for this reaction at 756 K.

Explanation:

$COCl_2\rightleftharpoons CO+Cl_2$

Equilibrium concentration of $COCl_2$

$[COCl_2]=7.40\times 10^{-4} M$

Equilibrium concentration of $CO$

$[CO]=3.76\times 10^{-2} M$

Equilibrium concentration of $Cl_2$

$[Cl_2]=1.78\times 10^{-4} M$

The expression of an equilibrium constant can be written as;

$K_c=\frac{[CO][Cl_2]}{[COCl_2]}$

$=\frac{3.76\times 10^{-2}\times 1.78\times 10^{-4}}{7.40\times 10^{-4}}$

$K_c=9.044\times 10^{-3}$

$9.044\times 10^{-3}$ is the value of the equilibrium constant for this reaction at 756 K.

5 0

3 years ago

Energy is the ability to do work or produce heat.<br> True<br> False

Dmitrij [34]

Answer: true

Explanation: hope this helpes

5 0

3 years ago

Read 2 more answers

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

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3 years ago

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Specify the steps included in the sedimentary rock formation

AlexFokin [52]

Answer:

Sedimentary rocks are the product of 1) weathering of preexisting rocks, 2) transport of the weathering products, 3) deposition of the material, followed by 4) compaction, and 5) cementation of the sediment to form a rock. The latter two steps are called lithification.

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3 years ago

A 2.5 mg sample of magnesium powder is ignited with 2 mg oxygen in a sealed container. All of the magnesium is consumed and 4.15

lys-0071 [83]

Answer:

Total mass of the reactant = 2+2.5 =4.5 mg

Total mass of product = 4.15 mg

therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g

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3 years ago