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Pavel [41]
3 years ago
6

Pb(NO3)2(aq) + KCl(aq) ---- → KNO3(aq) + PbC12(s)

Chemistry
1 answer:
emmasim [6.3K]3 years ago
6 0
If you’re asking to balance the equation then:

Pb(NO3)2(aq) + 2KCl(aq) -> 2KNO3(aq) + PbCl2(s)

Just remember: the equations at the end is Cl not C12

Note: the small number on the bottom (subscripts) apply to the one element if it’s inside the bracket and if the small number is on the outside of the bracket it applies to all the elements. For example the 3 in (NO3)2 applied only to the O (oxygen) and the 2 applies to both N and O but don’t forget it’s multiplied. So it would be 2 N’s and 6 O’s bc the 3 multiplies with the 2 only for the O.
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3 0
3 years ago
Read 2 more answers
Identify the sample and analyte in each of the scenarios.
MrMuchimi

Answer:

a) Analyte: lead. Sample: paint.

b) Analyte: nitrate. Sample: soil.

c) Analyte: citric acid. Sample: Lime

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Explanation:

A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.

Thus:

a) Analyte is lead while you must take a sample of paint to analyze this lead.

b) Analyte is the nitrate while sample must be soil.

c) Analyte is citric acid and lime is the sample

1) Lead: Analyte.

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7 0
3 years ago
a chunk of lead at 91.6 C was added to 200.0g of water at 15.5 C. The specific heat of lead is 0.129J/g C. When the temperature
worty [1.4K]
Let's apply the conservation of energy through the equation below:

Q for lead + Q for water = 0
So,
Q for lead = -Q for water
where
Q = mass*specific heat*(T₂ - T₁)
The specific heat of liquid water is 4.187 J/g·°C

Substituting the values:
(m)(0.129 J/g·°C)(17.9 - 91.6°C) = -(200 g)(4.187 J/g·°C)(17.9 - 15.5°C)
Solving for m,
m = <em>211.39 grams of lead</em>

4 0
3 years ago
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