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What is the relationship between the source region of an air mass and the properties of the air mass?

Levart [38]

A source region must have certain temperature and humidity properties that can remain fixed for a substantial length of time to affect air masses above it. above it. Air mass source regions occur only in the high or low latitudes; middle latitudes are too variable.

5 0

3 years ago

How much of strontium-90 will be left after 40 hours if you start with 960 grams and the half-life is 10 hours?

zmey [24]

Answer:

5 hours

Explanation:

7 0

2 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

dem82 [27]

Answer:

The percent yield of the reaction is 62.05 %

Explanation:

Step 1: Data given

Volume of methane = 25.5 L

Pressure of methane = 732 torr

Temperature = 25.0 °C = 298 K

Volume of water vapor = 22.0 L

Pressure of H2O = 704 torr

Temperature = 125 °C

The reaction produces 26.0 L of hydrogen gas measured at STP

Step 2: The balanced equation

CH4(g) + H2O(g) → CO(g) + 3H2(g)

Step 3: Calculate moles methane

p*V = n*R*T

⇒with p = the pressure of methane = 0.963158 atm

⇒with V = the volume of methane = 25.5 L

⇒with n = the moles of methane = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 298 K

n = (p*V) / (R*T)

n = (0.963158 * 25.5 ) / ( 0.08206 * 298)

n = 1.0044 moles

Step 4: Calculate moles H2O

p*V = n*R*T

⇒with p = the pressure of methane = 0.926316 atm

⇒with V = the volume of methane = 22.0 L

⇒with n = the moles of methane = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 398 K

n = (p*V) / (R*T)

n = (0.926316 * 22.0) / (0.08206 * 398)

n = 0.624 moles

Step 5: Calculate the limiting reactant

For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2

H2O is the limiting reactant. It will completely be consumed (0.624 moles).

Methane is in excess. There will react 0.624 moles. There will remain 1.0044 - 0.624 moles = 0.3804 moles methane

Step 6: Calculate moles hydrogen gas

For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2

For 0.624 moles H2O we'll have 3*0.624 = 1.872 moles

Step 9: Calculate volume of H2 at STP

1.0 mol at STP has a volume of 22.4 L

1.872 moles has a volume of 1.872 * 22.4 = 41.9 L

Step 10: Calculate the percent yield of the reaction

% yield = (actual yield / theoretical yield) * 100 %

% yield = ( 26.0 L / 41.9 L) *100 %

% yield = 62.05 %

The percent yield of the reaction is 62.05 %

6 0

3 years ago

Read 2 more answers

The electron configuration of Chromium is?

Zinaida [17]

Answer:

[Ar] 3d⁵ 4s¹ ..................

7 0

4 years ago

Question 26 Suppose a flask is filled with of and of . The following reaction becomes possible: The equilibrium constant for thi

Blizzard [7]

Answer:

0.36 M

Explanation:

There is some info missing. I think this is the complete question.

Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:

N₂(g) +O₂(g) ⇄ 2 NO(g)

The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.

Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:

[N₂] = 0.30 mol / 0.250 L = 1.2 M

[NO] = 0.70 mol / 0.250 L = 2.8 M

We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.

N₂(g) +O₂(g) ⇄ 2 NO(g)

I 1.2 0 2.8

C +x +x -2x

E 1.2+x x 2.8 - 2x

The equilibrium constant (K) is:

$K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}$

Solving for x, the positive one is x = 0.3601 M

[O₂] = 0.3601 M ≈ 0.36 M

7 0

3 years ago