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3 years ago

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What volume and mass of steam at 100. °C and 1.00 atm would release the same amount of energy during condensation as 100. cm^3 o

f liquid watet would release during freezing?

1 answer:

viva [34]3 years ago

4 0

Check on google and make it a question Maybe itll appear

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Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc

marysya [2.9K]

Explanation:

1) $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)$

$Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091$

2) $H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)$

$Kc=\frac{[H_2SO_4]}{[SO_3]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3$

3) $P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)$

$Kc=\frac{1}{[O_2]^3}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641$

5 0

3 years ago

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is $2.83\times 10^{22}$

<u>Explanation:</u>

<u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

<u>For b:</u>

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

2 years ago

7 organic compounds of plant origin

Nat2105 [25]

Explanation:

Among the numerous types of organic compounds, four major categories are found in all living things carbohydrates, lipids, proteins and nucleic acids.

8 0

3 years ago

How many atoms are equal to 1.5 moles of hellium

ANTONII [103]

Answer:

There are 1.8×1024 atoms in 1.5 mol HCl

Explanation:

6 0

2 years ago

Read 2 more answers

A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a

trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 + 2 HCl --> MnCl2 + H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl) X (0.0020 mole HCl / 1000ml) X (4.86 ml)

= 4.86 x 10^-3 mole

this is now dissolved in (70 + 4.86) = 74.86 ml or 0.07486 L

thus [Mn(OH)2] = 4.86 x 10^-3 mole / 0.07486 L = 0.064921 M

Ksp = [Mn2+][OH-]^2 = 4x^3 = 4(0.064921)^3 = 10.945 x 10^-4

6 0

3 years ago