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3 years ago

Click on the molecules that represent transpiration.

Chemistry

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Transpiration is the releasing of Water through Stomata
So, the molecule is H2O

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For a 0.300 mol sample of helium gas in a 0.200 L container at 248K, will the pressure be greater if calculated with the ideal g

konstantin123 [22]

Answer:

It changes by roughly 1 atm.

Explanation:

Hello!

In this case, since the ideal gas equation differs from the van der Waals' one by the presence of the a and b parameters which correct the assumption of no interactions into the container, they are written as:

$P=\frac{nRT}{V}\\\\P=\frac{RT}{v_m-b}-\frac{a}{v_m^2}$

Thus, the pressure via the ideal gas equation is:

$P=\frac{0.300mol*0.082\frac{atm*L}{mol*K}*248K}{0.200L}=30.5atm$

And the pressure via the van der Waals equation, considering the molar volume (vm=0.200L/0.300L=0.667L/mol) is:

$P=\frac{0.082\frac{atm*L}{mol*K}*248K}{0.667L/mol-0.0237L/mol}-\frac{0.0342atm*L^2/mol^2}{(0.667L/mol)^2}\\\\P=31.6atm-0.0769atm\\\\P=31.5atm$

It means that the pressure change by 1 atm, which is not a significant difference for helium.

5 0

3 years ago

Carbon Monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K, the equilibrium constant is 5.10. Calculate t

ozzi

Answer: Concentration of $CO_2$ at equilibrium= 1.386 M

Concentration of $H_2$ at equilibrium = 1.386 M

Concentration of $CO$ at equilibrium = 0.614 M

Concentration of $H_2O$ at equilibrium= 0.614 M

Explanation:

Moles of $CO$ = 1.00 mole

Moles of $H_2O$ = 1.00 mole

Moles of $CO_2$ = 1.00 mole

Moles of $H_2O$ = 1.00 mole

Volume of solution = 1.00 L

Initial concentration of $CO$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $H_2O$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $CO_2$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $H_2$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2$

Initial conc. 1.00M 1.00 M 1.00 M 1.00 M

At eqm. conc. (1.00-x) M (1.00-x) M (1.00+x) M (1.00+x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}$

Now put all the given values in this expression, we get :

$5.10=\frac{(1.00+x)^2}{(1.00-x)^2}$

By solving the term 'x', we get :

x = 0.386

Concentration of $CO_2$ at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of $H_2$ = (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of $CO$ = (1.00-x) M = (1.00-0.386) M = 0.614 M

Concentration of $H_2O$ = (1.00-x) M = (1.00-0.386) M = 0.614 M

3 0

3 years ago

Irina-Kira [14]

The correct answer is D

6 0

3 years ago

Consider the following enthalpy diagram and enthalpies of intermediate and overall chemical reactions.

SashulF [63]

Answer: The overall chemical reaction is exothermic is the correct statement.

Explanation: As seen from the given image, the reactants are at higher energy level and products are at lower energy level and the excess energy is released in the form of heat. These reactions are considered as exothermic reactions.

Hence, the reaction given in the image is an exothermic reaction.

6 0

3 years ago

Read 2 more answers

Calculate the percent by mass of 4.35g of Na I dissolved in 105g of water

Ludmilka [50]

<u>We are given:</u>

Mass of Na added = 4.35 grams

Mass of water = 105 grams

<u>Mass Percent of Na:</u>

Total mass of the solution = mass of solute + mass of solvent

Total mass of the solution = 4.35 + 105 = 109.35 grams

Mass percent of solute = (mass of solute / mass of solution) * 100

Mass percent of Solute = (4.35 / 109.35) * 100

Mass percent = 3.978 %

4 0

3 years ago