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lesya692 [45]
1 year ago
14

An unknown quantity of methane gas, CHA, is held in a 2.00-liter container at 77°C. The pressure inside the container is 3.00 am

. How many moles of methane must be in the container?
Chemistry
1 answer:
kolbaska11 [484]1 year ago
6 0

Answer:

0.21\text{ mol}

Explanation:

Here, we want to calculate the number of moles of methane in the container

From the ideal gas law:

\begin{gathered} PV\text{ = nRT} \\ n\text{ = }\frac{PV}{RT} \end{gathered}

where:

P is the pressure inside the container which is 3 atm

V is the volume of the container which is 2 L

R is the molar gas constant which is 0.0821 Latm/mol.k

T is the temperature in Kelvin (we convert the temperature in Celsius by adding 273 : 273 + 77 = 350 K)

n is the number of moles that we want to calculate

Substituting the values, we have it that:

n=\text{ }\frac{3\times2}{0.0821\text{ }\times350}\text{ = 0.21 mol}

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The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g
ANTONII [103]

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

7 0
4 years ago
A new metal alloy is found to have a specific heat capacity of 0.260 J/(g⋅∘C). First, 47 g of the new alloy is heated to 180. ∘C
Vlad1618 [11]

Answer:

The final temperature, at the equilibrium is 24.14 °C

Explanation:

Step 1: Data given

specific heat capacity of alloy = 0.260 J/(g°C)

MAss of alloy = 47 grams

Mass of water = 110 grams

Specific heat of water = 4.184 J/g°C

Initial temperature of water = 20.0 °C

Initial temperature of alloy = 180.0 °C

Step 2: Calculate the final temperature at equilibrium

Heat lost = heat gained

Qlost = -Qgained

Q(alloy) =- Q(water)

Q=m*c*ΔT

Q = m(alloy)*c(alloy)*ΔT(alloy) = -m(water) * c(water)* ΔT(water)

⇒with m(alloy) = the mass of alloy = 47.0 grams

⇒with c(alloy) = the specific heat of alloy = 0.260 J/g°C

⇒with ΔT(alloy) = the change of temperature = T2- T1 = T2 - 180 °C

⇒with m(water) = the mass of water = 110 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature = T2 - 20.0°C

47.0*0.260 * (T2 - 180.0) = - 110 * 4.184 * (T2 - 20.0)

12.22(T2-180.0) = -460.24(T2- 20)

12.22T2 - 2199.6 = -460.24T2 + 9204.8

472.46T2 = 11404.4

T2 = 24.14 °C

The final temperature, at the equilibrium is 24.14 °C

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