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frez [133]
2 years ago
11

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

01 kpa. if the molecular weight of air is 29 g/mol, what is the mass of air inside the refrigerator? the ideal gas constant is r = 8.314 j/mol ∙ k = 0.0821 l ∙ atm/mol ∙ k.
Chemistry
1 answer:
Dafna11 [192]2 years ago
6 0
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 
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Explanation:

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BaSO4 contains four oxygen atoms.

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Discuss some of the biotic (living) and abiotic (nonliving) factors in the chimps’ ecosystem that affect their behavior.
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Explanation:

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2 years ago
How and why do ionic bonds form? Check all of the boxes that apply. Ionic bonds form between metal atoms and other metal atoms.
sammy [17]

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8 0
2 years ago
Read 2 more answers
Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

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7 0
2 years ago
The volume of a single strontium atom is 4.15×10-23 cm3. What is the volume of a strontium atom in microliters
Ivenika [448]

Answer:-  4.15*10^-^2^0\mu L

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We know that:

1cm^3 = 1 mL

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and, 1L=10^6\mu L

Let's use these conversions factors for the desired conversion using dimensional as:

4.15*10^-^2^3cm^3(\frac{1mL}{1cm^3})(\frac{10^-^3L}{1mL})(\frac{10^6\mu L}{1L})

= 4.15*10^-^2^0\mu L

So, the answer is  4.15*10^-^2^0\mu L .

7 0
1 year ago
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