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ss7ja [257]
3 years ago
12

A student has two objects. Object 1 has a mass of 28 g and a volume of 12 cubic

Chemistry
1 answer:
bearhunter [10]3 years ago
8 0

Answer:

Object 2 with a density of 0.68

Explanation:

object 2 has a lower density than water so it will float

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328. mL of 0.00345 M NaI (aq) is combined with 703. mL of 0.00802 M Pb(NO3)2 (aq). Determine if a precipitate will form given th
Vsevolod [243]

Answer:

The solution will not form a precipitate.

Explanation:

The Ksp of PbI₂ is:

PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)

Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>

When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:

[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>

[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>

<em />

Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>

If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.

Replacing:

Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹

As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.

6 0
3 years ago
A balloon inflated in a room at 27 degrees celsius has a volume of 8.00 L.the balloon is then heated to a temperature of 78 degr
dangina [55]
This problem is being solved using Ideal Gas Equation.
                                                   PV  =  nRT
Data Given:
                  Initial Temperature = T₁ = 27 °C = 300 K
                  Initial Pressure      =  P₁ = constant
                  Initial Volume         = V₁ = 8 L
                   Final Temperature = T₂ = 78 °C = 351 K
                  Final Pressure      =  P₂ = constant
                  Final Volume         = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
                                          V₁ / T₁  =  V₂ / T₂
Solving for V₂,
                                           V₂  =  (V₁ × T₂) ÷ T₁
Putting Values,
                                           V₂  =  (8 L × 351 K) ÷ 300 K

                                           V₂  =  9.38 L
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