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snow_tiger [21]
2 years ago
8

Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Chemistry
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

Mass_{chromium}=52.1\ g

Mass_{neon}=8.23\times 10^{-7}\ g

Explanation:

<u>Calculation of the mass of chromium as:- </u>

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

1.002\ mol= \frac{Mass}{51.9961\ g/mol}

Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g

<u>Calculation of the mass of neon as:- </u>

Moles = 4.08\times 10^{-8} moles

Molar mass of neon = 20.1797 g/mol

Thus,

1.002\ mol= \frac{Mass}{20.1797\ g/mol}

Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g

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The interior of a refrigerator has a volume of 0.600 m3. the temperature inside the refrigerator in 282 k, and the pressure is 1
Dafna11 [192]
In this instance we can use the ideal gas law equation to find the number of moles of gas inside the refrigerator 
PV = nRT
where 
P - pressure - 101 000 Pa
V - volume - 0.600 m³
n - number of moles
R - universal gas constant - 8.314 J/mol.K
T - temperature - 282 K
substituting these values in the equation 
101 000 Pa x 0.600 m³ = n x  8.314 J/mol.K x 282 K
n = 25.8 mol
there are 25.8 mol of the gas
to find the mass of gas
mass of gas = number of moles x molar mass of gas
mass = 25.8 mol x 29 g/mol  = 748.2 g
mass of gas present is 748.2 g 
6 0
1 year ago
How much heat is released or absorbed (in kJ) by the system in the reaction of 17.7 g of SF6 with 23.7 g of H2O?
Jlenok [28]

Answer:

+279.744 kJ.

Below is an attachment containing the solution to question.

5 0
2 years ago
A scientist performs an experiment in which they create an artificial cell with a selectively permeable membrane through which o
lozanna [386]

Answer:

Water moves into the cell

Explanation:

As shown in the question above, the cell is high in glucose and placed in a glass filled with water. This cell has a semi permeable membrane that allows only water to pass through, as the concentration of water within the cell is low, the cell will attempt to strike a balance with the medium it is inserted into. For this reason, what is likely to happen is the passage of water from the most concentrated to the least concentrated medium, that is, the water will pass from the cup to the cell.

water moves into the cell through osmosis.during osmosis water moves from a region of low concentration of solute to a region of high concentration of solute.the glucose introduced into the cell makes it more concentrated.

In this case the cell is hypertonic and water would enter into the cell through the semi permeable membrane.this membrane allows water to pass through but not glucose.this movement of water into the cell causes the cell to become turgid.

8 0
1 year ago
A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9
Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂)  = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺)   = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻)    = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

      = -260.75 + 232.34

     = -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

   = 1/(0.750 * 0.232²)

  = 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

     = -28.41 +7.95

    = -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

8 0
1 year ago
Deuterium (D or 2H) is an isotope of hydrogen. The molecule D2 undergoes an exchange reaction with ordinary hydrogen, H2, that l
Paladinen [302]

Answer:

Kp_2} =2.0

Explanation:

K_{p1}= 1.8

K_{p2}= ???

T_1= 298K

T_2 = 415 K

\delta H = 0.64 kJ/mol = 640 J/mol

R = 8.314 J/mol.K

Using Van't Hoff Equation:

In (\frac{Kp_2}{Kp_1} )=(-\frac{DH}{R} *(\frac{1}{T_2}-\frac{1}{T_1}  )

In (\frac{Kp_2}{1.8} )=(-\frac{640}{8.314} *(\frac{1}{415}-\frac{1}{298}  )

In (\frac{Kp_2}{1.8} )=0.072827

(\frac{Kp_2}{1.8} )= e^{0.0728727}

\frac{Kp_2}{1.8} =1.075593605

Kp_2} =1.075593605*1.8

Kp_2} =1.936068489

Kp_2} =2.0

3 0
2 years ago
Read 2 more answers
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