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kodGreya [7K]
3 years ago
6

A sample of gas occupies 9.0 mL at a pressure of 500.0 mm Hg. A new volume of the same sample is at a pressure of 750.0 mm Hg. I

s the new pressure smaller or larger than the original? Therefore, is the new volume larger or smaller than 9.0 mL? What is this new volume?
Chemistry
2 answers:
cluponka [151]3 years ago
4 0

Answer: The new pressure of 750 mmHg is greater than initial pressure of 500 mm Hg and the new volume of 6.0 ml is lesser than the initial volume of 9.0 ml.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 500.0 mmHg

P_2 = final pressure of gas= 750.0 mmHg

V_1 = initial volume of gas = 9.0 ml

V_2 = final volume of gas  = ?

500\times 9.0=750\times V_2  

V_2=6.0ml

Therefore, the gas would compress to new volume of 6.0 ml.

The new pressure of 750 mmHg is greater than initial pressure of 500 mm Hg and the new volume of 6.0 ml is lesser than the initial volume of 9.0 ml.

Ymorist [56]3 years ago
3 0
The new pressure is larger than the original, the new volume is smaller than 9.0 ml and the new volume is 6.0 

good luck :D
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Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The
Maksim231197 [3]

Answer:

           \large\boxed{\large\boxed{24.6kJ}}

Explanation:

<u>1. Energy to heat the liquid water from 55ºC to 100ºC</u>

     Q=m\times C\times \Delta T

  • m = 10.1g
  • C = 4.18g/JºC
  • ΔT = 100ºC - 55ºC = 45ºC

     Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J

<u>2. Energy to change the liquid to steam at 100ºC</u>

      L=\lambda \times n

  • λ = 40.6kJ/mol
  • n = 10.1g / 18.015g/mol = 0.5606mol

      L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J

<u>3. Total energy</u>

       1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ

7 0
3 years ago
The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water
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PbSO₄ partially dissociates in water. the balanced equation is;
                    
                       PbSO₄(s) ⇄  Pb²⁺(aq) + SO₄²⁻(aq)
Initial                                     -                -
Change             -X               +X           +X
Equilibrium                           X              X

Ksp           =    [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸  =    X * X
1.6 x 10⁻⁸  =    X²
          X    =   1.3 x 10⁻⁴ M
      
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M. 
[Pb²⁺]  = 1.3 x 10⁻⁴ M.
           = 1.3 x 10⁻⁴ mol / L x 207 g / mol 
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Answer:

B = (2.953 × 10⁻⁹⁵) N.m⁹

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At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.

That is,

Fa + Fr = 0

Fa = - Fr

Fa = (|q₁q₂|)/(4πε₀r²)

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Fr = -B/r⁹

(|q₁q₂|)/(4πε₀r²) = (B/r⁹)

|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C

(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²

r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m

(k|q₁q₂|)/(r²) = (B/r⁹)

(k × |q₁q₂|) = (B/r⁷)

B = (k × |q₁q₂| × r⁷)

B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]

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