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kozerog [31]
3 years ago
14

Unit 1- Unit Assessment (Matter & Density)

Chemistry
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:

<h3>The answer is 11 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 3025 g

volume = 275 mL

We have

density =  \frac{3025}{275}  \\

We have the final answer as

<h3>11 g/mL</h3>

Hope this helps you

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When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
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Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

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n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

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Explanation:

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05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises
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Answer:

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Explanation:

The following data were obtained from the question:

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Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

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Next, we shall determine the volume of the gold. This can be obtained as follow:

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Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

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Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

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