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mariarad [96]
3 years ago
7

Compare melting point freezing point and boiling point

Chemistry
1 answer:
Tresset [83]3 years ago
4 0
Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase
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8. How much enthalpy/heat is transferred when 0.5113
Gennadij [26K]
First convert to moles:
0.5113 g / 17 g/mol = 0.0301 mol

Now create a ratio based on the reaction provided to solve for the unknown:

4 NH3 / -905.4 kJ = 0.0301 mol NH3 / x kJ
x = -6.808 kJ
8 0
3 years ago
Molecular iodine, I2(g), dissociates into iodine atoms at 625 K with a first-order rate constant of 0.271 s−1 .
katrin [286]
The correct is this because they said
4 0
2 years ago
SA
inn [45]

Answer:

маликулярная масса ровна  23

8 0
3 years ago
2NaCl → 2Na +Cl2<br> What reaction is this
babunello [35]

Answer:

Decomposition reaction

Explanation:

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5 0
2 years ago
Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
3 years ago
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