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Inessa05 [86]
3 years ago
13

2KI(aq) + Pb(NO,)(aq) › PbI(s) + 2KNOg(aq) Balance the equation

Chemistry
1 answer:
azamat3 years ago
5 0

Answer:

Explanation:

There are some mistakes in writing the equation. However, Please check the below one and contact me again if necessary.

2KI(aq) +  Pb(NO₃)₂ (aq) -> PbI₂ (s) + 2KNO₃

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<span>There are 1000 cm3 in 1 liters. Hence 1 liter of the liquid would weigh: 1000 cm3 x (1.17 g/cm3) = 1170 gm and there are 1000 gm in 1 kg, so we want enough liters to have a mass of 3.75 kg x 1000 gm/kg = 3750 gm Hence, # of liters = desired mass / # of gm per liter = 3750 gm / 1170 gm/liter = 3.2051282 liters</span>
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Which of the following would be most likely to experience strong intermolecular forces?
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The correct answer is D

Explanation:

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Do scientists worldwide use the same units of measurements?
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The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,
Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by t_{\frac{1}{2}

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. \frac{2400}{2}=1200, after second  half life, the activity would be reduced to half of 1200 i.e. \frac{1200}{2}=600,  and after third half life, the activity would be reduced to half of 600 i.e. \frac{600}{2}=300,

Thus the activity 9 minutes later is 300 bq.

7 0
3 years ago
9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0
3 years ago
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