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3 years ago

How many grams of water vapor (H2O) are in a 10.2 liter sample of 0.98 atmosphere and 26 C

1 answer:

5 0

Hey there!

<span>Use the equation of Clapeyron:
</span>
T in kelvin :

26 + 273.15 => 299.15 K

R = 0.082

V = 10.2 L

P = 0.98 atm

number of moles :

P *V = n * R * T

0.98 * 10.2 = n * 0.082 * 299.15

9.996 = n * 24.5303

n = 9.996 / 24.5303

n = 0.4074 moles

Therefore:

Molar mass H2O = 18.01 g/mol

1 mole H2O ------------- 18.01 g
0.4074 moles ----------- m

m = 0.4074 * 18.01 / 1

m = 7.339 g of H2O

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A gas has a volume of 490. mL at a temperature of -35.0 degrees C. What volume would the gas occupy at 42.0 degrees Celsius? Ple

miskamm [114]

Answer:

648.5 mL

Explanation:

Here we will assume that the pressure of the gas is constant, since it is not given or specified.

Therefore, we can use Charle's law, which states that:

"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically:

$\frac{V}{T}=const.$

where

V is the volume of the gas

T is its absolute temperature

The equation can be rewritten as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have:

$V_1=490 mL$ is the initial volume of the gas

$T_1=-35.0^{\circ} + 273 = 238 K$ is the initial temperature

$T_2=42.0^{\circ}+273=315 K$ is the final temperature

Solving for V2, we find the final volume of the gas:

$V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL$

8 0

3 years ago

Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

According to table g, which substance forms an unsaturated solution when 80. grams of the substance are stirred into 100. grams

e-lub [12.9K]

A saturated solution is one in which no more solute is able to dissolve in a given solvent at a particular temperature. Some amount of the solute is left undissolved in the solution.

Unsaturated solution has solute in lower proportions than required to form a saturated solution.

Supersaturated solution has solute in amounts greater than a saturated solution.

We can take the help of solubility curve in order to find out the amount of a salt required to prepare a saturated solution of that salt at a particular temperature.

The solubility of KI at 10 $^{0}C$ is 136 g/ 100 mL water

The solubility of $KNO_{3}$ at $10^{0}C$ is 21 g/100 mL water.

The solubility of $NaNO_{3}$ at $10^{0}C$ is 80 g/100 mL water.

The solubility of NaCl at $10^{0}C$ is 38 g/ 100 mL water.

So the correct answer will be KI, as it would need 136 g KI / 100 mL water to form a saturated solution at $10^{0}C$ .So, if we have 80g KI/ 100mL water it would be an unsaturated solution.

4 0

3 years ago

Read 2 more answers

Are the following combinations allowed? If not, show two ways to correct them:

mafiozo [28]

The following combination of n=3 ; l=1 ; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

<h3>Is the following combination n=3; l=1; ml=-2 allowed or not.? If not, suggest two ways through which it can be corrected.</h3>

The following combination of n=3 ; l=1 ; ml=-2 is not allowed.

There are several rules that need to be followed for assigning electron quantum numbers. They are:

1. Principal quantum number should be 1 ≤ n

2. Azimuthal quantum number, 0 ≤ l ≤ n − 1

3. Magnetic quantum number, -l ≤ ml ≤ l

4. Spin quantum number as either -1/2 or +1/2

For n = 3,

l should be n - 1 or n - 2 or n - 3 = 2, 1, 0 respectively.

If we choose l = 1 then ml should be -1, 0 and +1

Therefore, one way to correct the combination would be to change the magnetic quantum number to -1

If we choose l = 2 then ml would be -2, -1, 0, +1, +2

Thus, another way to correct the combination is to choose the azimuthal quantum number as 2.

Thus, the following combination of n=3; l=1; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

To learn more about quantum numbers refer:

brainly.com/question/5927165

#SPJ4

5 0

1 year ago

Find the kinetic energy of a boy of mass 6.5kg running at a velocity of 6.0m/s

ololo11 [35]

Explanation:

Given,

M=6.5kg,velocity(v)=6m/s

K.E=1/2mv^2

=1/2×6.5×6^2

=1/2×6.5×36

=18×6.5

=117 J

3 0

2 years ago