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Sindrei [870]
3 years ago
15

Calculate the mass of 1.00 mol of each of these substances

Chemistry
1 answer:
bogdanovich [222]3 years ago
7 0
What are the substances?
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HElp i am stuck and i am not sure
alexandr1967 [171]

Answer:

B

Explanation:

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7 0
2 years ago
An alpha particle is equivalent to a _____ nucleus.
astra-53 [7]
Alpha particle is equivalent to B. Helium atom (2 protons, 2 neutrons) 
6 0
3 years ago
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What is the constant parameters of pressure and volume? The relationship between the two? Are they directly proportional or inve
OverLord2011 [107]
Now I'm just going to assume you mean Charles law. So when working with gases, there are 4 properties: pressure, volume, temp, and quantity. The simple gas laws deal with 2, while leaving the other 2 constant. If Charles' Law changes temp and volume, what 2 stay constant? Pressure and quantity
6 0
3 years ago
In general, in what type of solvent (non-polar, moderately polar, or highly polar) are polar solutes most soluble? Explain why.
tia_tia [17]

Answer:

  • In general, polar solutes are most soluble in highly polar solvents.

Explanation:

The general rule is "like dissolves like" which means that <em>polar solvents </em>dissolve polar (or ionic) <em>solutes</em> and <em>non-polar solvents</em> dissolve non-polar solutes.

In order for a solvent dissolve a solute, the strength of the interacttion (force) between the solute and the solvent units (atoms, molecules, or ions) must be stronger than the strength of the forces that keep together he particles of the pure substances (known as intermolecular forces).

Since the nature of the interactions between the units are electrostatic, the more polar is the solvent the better it will be able to attract and surround the solute particles, keeping them separated and in solution. That mechanism explains why polar solutes will be most soluble in highly polar solvents.

5 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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