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antoniya [11.8K]
1 year ago
8

Write a reaction and balance it for the reaction of lead with hydrochloric acid

Chemistry
1 answer:
LenKa [72]1 year ago
8 0

A balanced the reaction of lead with hydrochloric acid:

Pb(aq) + 2HLC (l) → PbCl2(aq) + H2(g)

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If an ice cube weighing 25.0 g with an initial
riadik2000 [5.3K]

Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

3 0
2 years ago
A galvanic^cell consists of left ompartmcnt with a tin elcctrode in contact with 0.1 M Sn(NO_3)_2(aq) and a right compartment wi
lana66690 [7]

Answer:

Electrons will flow from left to right through the wire.

Pb^2+ ions will be reduccd to Pb metal.

The concentration of Sn2+ ions in the left compartment will increase.

Explanation:

Looking at the relative electrode potentials of the two metals

Sn= -0.14

Pb=-0.13

Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.

6 0
2 years ago
How are protons, electrons, and neutrons correctly compared​
Fiesta28 [93]

they all have one thing in common and that its all made up of atoms. When these components are active it creates energy  

4 0
2 years ago
what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all
Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

#SPJ4

5 0
10 months ago
Nitric oxide (NO) reacts readily with chlorine gas as follows.2 NO(g) + Cl2(g) equilibrium reaction arrow 2 NOCl(g)At 700. K the
Yanka [14]

<u>Answer:</u>

<u>For a:</u> The mixture will need to produce more reactants to reach equilibrium.

<u>For b:</u> The mixture will need to produce more reactants to reach equilibrium.

<u>For c:</u> The mixture will need to produce more products to reach equilibrium.

<u>Explanation</u>:

For the given chemical equation:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

The expression of K_{p} for above equation follows:

K_{p}=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}   .....(1)

We are given:

Value of K_p = 0.26

There are 3 conditions:

  • When K_{p}>Q_p; the reaction is product favored.
  • When K_{p}; the reaction is reactant favored.
  • When K_{p}=Q_p; the reaction is in equilibrium.

For the given options:

  • <u>For a:</u>

We are given:

p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm

Putting values in expression 1, we get:

Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57

As, K_{p}; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

  • <u>For b:</u>

We are given:

p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm

Putting values in expression 1, we get:

Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6

As, K_{p}; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

  • <u>For c:</u>

We are given:

p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm

Putting values in expression 1, we get:

Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008

As, K_{p}>Q_p; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.

8 0
3 years ago
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