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Murljashka [212]
3 years ago
7

3. Give two examples of an ionic compound.

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
5 0

Answer: NaCl and LiNO3

Explanation:

Because in aqueous solution they ionize

NaCl produces Na+ and Cl-

And LiNO3 produces Li+ and NO3- ion

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The diagram below shows a cell placed in a solution.
Sholpan [36]

it will expand as water moves into it.


3 0
3 years ago
Newton's third law states that for every action, there is an ______ and ____ reaction force and that forces come in ____
Lelechka [254]

Answer:

Newton's third law of motion states that every action, there is an equal and opposite reaction force and that forces come in pairs

3 0
2 years ago
D5W is a solution used as an intervenous fluid.It is 5.0% by mass solution of the dextrose (C6H12O6) in water. if the density of
Alona [7]

Answer:

0.29 mol/L

Explanation:

Its density is 1.029 g/ml so in a liter (1000 mL) there is 1029 g of solution, but only 5% is dextrose.

0.05x1029=51.45

So in a liter of D5W solution there is 51.45 g of dextrose.

Dextrose molar mass iss 180.156 g/mol, so in 51.45 g of dextrose there is

51.45/180.156=0.29 mol

In one liter of solution there is 0.29 mol of dextrose, so the molarity of such solution is 0.29 mol/L.

6 0
3 years ago
Show your work with good use of units, rounding, and significant figures. [Hint: it is good practice to show the value of your a
mel-nik [20]

Heat required : 4.8 kJ

<h3>Further explanation </h3>

The heat to change the phase can be formulated :

Q = mLf (melting/freezing)

Q = mLv (vaporization/condensation)

Lf=latent heat of fusion

Lv=latent heat of vaporization

The heat needed to raise the temperature

Q = m . c . Δt

1. heat to raise temperature from -20 °C to 0 °C

\tt Q=10\times 2.09\times (0-(-20)=418~J

2. phase change(ice to water)

\tt Q=10\times 333=3330~J

3. heat to raise temperature from 0 °C to 25 °C

\tt Q=10\times 4.18\times (25-0)=1045~J

\tt Q~tot=418+3330+1045=4793~J\rightarrow rounding~and~2~sig~figs=4.8~kJ

3 0
2 years ago
How much heat, in calories, is needed to raise the temperature of 125.0 g of Lead (c lead = 0.130 J/g°C) from 17.5°C to 41.1°C?
pav-90 [236]
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J

(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
are needed.
8 0
3 years ago
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