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Anon25 [30]
3 years ago
10

Which term refers to compounds that can form hydrates but do not contain water molecules?

Chemistry
2 answers:
Lerok [7]3 years ago
6 0
The term that refers to compounds that can form hydrates but do not contain water molecules is anhydrous.
babunello [35]3 years ago
6 0

A or 1st option: anhydrous

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Explain what might happen to tundra animals, such as polar bears, as earth's climate warms. pls explain
SCORPION-xisa [38]

Answer:

They will die.

Explanation:

Polar Bears were made to live in the frigid temperatures of the tundras. If there is a temperature change it can affect their habitat. It can destroy it and they would have to move out to another place. They can die due to global warming.

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2 years ago
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3 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

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3 years ago
How does a scientist explain something when a controlled experiment cannot be carried out
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7 0
3 years ago
You burn a 15g jellybean to warm 50 mL of water, which increases the temperature of the water 25 °C. How many calories of heat a
Juli2301 [7.4K]

Answer:

Q = 375\,cal

Explanation:

The quantity of heat transfered from the jellybean to the water is:

Q = \rho\cdot V \cdot c\cdot \Delta T

Q = \left(1\,\frac{g}{cm^{3}}\right)\cdot (15\,cm^{3})\cdot \left(1\,\frac{cal}{g\cdot ^{\circ} C} \right)\cdot (25\,^{\circ}C )

Q = 375\,cal

7 0
2 years ago
Read 2 more answers
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