Answer:
Option A. 107 mL
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 150 mL
Initial pressure (P₁) = 500 mmHg
Final pressure (P₂) = 700 mmHg
Temperature = constant
Final volume (V₂) =?
The final volume of the gas can be obtained by using the Boyle's law equation as shown below:
P₁V₁ = P₂V₂
500 × 150 = 700 × V₂
75000 = 700 × V₂
Divide both side by 700
V₂ = 75000 / 700
V₂ = 107 mL
Therefore, the final volume of the gas is 107 mL.
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Answer:
Ok; just use PV=nRT solve for n.
Explanation:
Answer:
Heat going into a substance changes it from a solid to a liquid or a liquid to a gas. Removing heat from a substance changes a gas to a liquid or a liquid to a solid.
Liquid → Gas:
VaporizationGas → Liquid:
CondensationSolid → Liquid:
Melting or fusion
Solid → Gas: Sublimation
Explanation:
The given question tells you that a certain piece of wire has a mass = 2.0 g per meter
That means that if you consider a piece of wire that is 1 m in length, its mass will be equal to 2.00 g .
According to question,
Now, you know that
1 m = 100 cm
Mass of 2.00 g of copper will correspond to a wire that is a 100 cm long.
This implies that 0.28 gof copper will correspond to a wire that is
0.28 g * 100 cm / 2.0 g = 14 cm long
Hence, 14 cm of the wire would be needed to provide 0.28 g of copper.
Properties of Copper :
High conductivity and ductility.
Non magnetic.
To know more about Copper here:
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