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3 years ago

Calculate the volume of 2.65 moles of CO2 gas at STP.

Chemistry

1 answer:

Harman [31]3 years ago

6 0

Answer:

The volume is 59, 3 liters. See the explanation below, please

Explanation:

STP conditions (standard) correspond to 273K of temperature and 1 atm of pressure. These values are used and the volume is calculated, according to the formula:

PV = nRT

1 atm x V= 2, 65 moles x 0, 082 l atm/K mol x 272 K

V= 2, 65 moles x 0, 082 l atm/K mol x 272 K/1 atm = 59, 3 liters

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A mixture of 0.682 mol of H2 and 0.440 mol of Br2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 70

ahrayia [7]

Answer: 0.294 mol of $Br_2$ present in the reaction vessel.

Explanation:

Initial moles of $H_2$ = 0.682 mole

Initial moles of $Br_2$ = 0.440 mole

Volume of container = 2.00 L

Initial concentration of $H_2=\frac{moles}{volume}=\frac{0.682moles}{2.00L}=0.341M$

Initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.440moles}{2.00L}=0.220M$

equilibrium concentration of $H_2=\frac{moles}{volume}=\frac{0.536mole}{2.00L}=0.268M$

The given balanced equilibrium reaction is,

$H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)$

Initial conc. 0.341 M 0.220 M 0 M

At eqm. conc. (0.341-x) M (0.220-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[HBr]^2}{[Br_2]\times [H_2]}$

we are given : (0.341-x) = 0.268 M

x= 0.073 M

Thus equilibrium concentration of $Br_2$ = (0.220-x) M = (0.220-0.073) M = 0.147 M

$[Br_2]=\frac{moles}{volume}\\0.147=\frac{xmole}{2.00L}\\\\x=0.294 mole$

Thus there are 0.294 mol of $Br_2$ present in the reaction vessel.

7 0

3 years ago

What will be the theoretical yield of tungsten(is) ,W, if 45.0 g of WO3 combines as completely as possible with 1.50 g of H2

Anarel [89]

Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃ + 3H₂ → 3H₂O + W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten(VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass . mol)

0.194 mol . 183.84 g/mol = 35.6 g

3 0

3 years ago

Why do people want different kinds of light bulbs?

ololo11 [35]

Answer:

LED bulbs fit standard light sockets and are the most energy-efficient option. LEDs have lower wattage than incandescent bulbs but emit the same light output. This allows them to produce the same amount of light but use less energy. LEDs can last over 20 years and don't contain mercury

7 0

3 years ago

10.

Andru [333]

Answer: 1.8 moles Fe and 2.7 moles $CO_2$ are produced.

Explanation:

The balanced chemical reaction is:

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

According to stoichiometry :

3 moles of $CO$ require 1 mole of $Fe_2O_3$

Thus 2.7 moles of $CO$ will require= $\frac{1}{3}\times 2.7=0.9moles$ of $Fe_2O_3$

Thus $CO$ is the limiting reagent as it limits the formation of product and $Fe_2O_3$ is the excess reagent.

As 3 moles of $CO$ give = 2 moles of $Fe$ and 3 moles of $CO_2$

Thus 2.7 moles of $CO$ will give = $\frac{2}{3}\times 2.7=1.8moles$ of $Fe$ and $\frac{3}{3}\times 2.7=2.7moles$ of $CO_2$

Thus 1.8 moles Fe and 2.7 moles $CO_2$ are produced.

7 0

3 years ago

What is the scientific method 1-7?

IRINA_888 [86]

#1- Identify a problem
#2- Collect info on your problem
#3- Make a hypothesis
#4- Design an experiment to test your hypothesis
#5- Collect data and observations
#6- Accept or reject your hypothesis
#7- Record results
Hope this helps.

3 0

4 years ago