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sergeinik [125]

4 years ago

13

Intermolecular forces exist between what?

1 answer:

Paraphin [41]4 years ago

6 0

Answer:

Intramolecular forces are the forces that hold atoms together within a molecule. Intermolecular forces are forces that exist between molecules.

Explanation:

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What is true about equinox?

Leno4ka [110]

Answer:

In technical astronomical terms, the equinox is the moment at which the sun appears to cross the equator as a result of Earth’ s rotation around the sun. The vernal (spring) equinox, which occurs as the sun moves from south to north across the equator, takes place around March 21 and marks the beginning of spring.

Explanation:

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3 years ago

Which hydrocarbon name adheres correctly to the IUPAC naming system?

tino4ka555 [31]

The hydrocarbon name that adheres correctly to the IUPAC naming system is 3-ethyl-2,4-dimethylhexane.

To add, <span>in chemical </span>nomenclature, the IUPAC nomenclatureof organic chemistry is a systematic method ofnaming<span> organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (</span>IUPAC).

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4 years ago

Which lists imaging techniques that use wave behaviors in order of increased risk of cancer from least risk to

gulaghasi [49]

Answer:

sonography, X-ray radiography, CT scan

Explanation:

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4 years ago

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Helen [10]

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3 years ago

Read 2 more answers

The following data is given to find the formula of a Hydrate:

Darya [45]

Answer:

$m_{CaSO_4.H_2O}=2.49\ g$

$m_{CaSO_4}=1.51\ g$

$m_{water}=0.98\ g$

$\%\ of\ water=39.36\ \%$

$moles_{water}= 0.0544\ mol$

$moles_{CaSO_4}= 0.0111\ mol$

$CaSO_4:H_2O$ = 5 : 1

Explanation:

Given :

$m_{crucible}=13.56\ g$

$m_{crucible}+m_{CaSO_4.H_2O}=16.05\ g$

$m_{crucible}+m_{CaSO_4}=15.07\ g$

<u>Mass of salt hydrate:</u>

$m_{crucible}=13.56\ g$

$m_{crucible}+m_{CaSO_4.H_2O}=16.05\ g$

$m_{CaSO_4.H_2O}=16.05-m_{crucible}\ g=16.05-13.56\ g=2.49\ g$

<u>Mass of salt anhydrous:</u>

$m_{crucible}=13.56\ g$

$m_{crucible}+m_{CaSO_4}=15.07\ g$

$m_{CaSO_4}=15.07-m_{crucible}\ g=15.07-13.56\ g=1.51\ g$

<u>Mass of water:</u>

$m_{water}=m_{CaSO_4.H_2O}-m_{CaSO_4}=2.49-1.51\ g=0.98\ g$

$m_{water}=2.49-1.51\ g=0.98\ g$

<u>Percentage of water:</u>

$\%\ of\ water=\frac{Mass_{water}}{Total\ mass\ of\ hydrated\ salt}\times 100$

$\%\ of\ water=\frac{0.98}{2.49}\times 100$

$\%\ of\ water=39.36\ \%$

<u>Moles of water:</u>

Mass of water = 0.98 g

Molar mass of $H_2O$ = 18 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{0.98\ g}{18\ g/mol}$

$moles_{water}= 0.0544\ mol$

<u>Moles of anhydrate salt:</u>

Amount = 1.51 g

Molar mass of $CaSO_4$ = 136.14 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles are:

$moles= \frac{1.51\ g}{136.14\ g/mol}$

$moles_{CaSO_4}= 0.0111\ mol$

The simplest ration of the two are:

$CaSO_4:H_2O$ = 0.0111 : 0.0544 = 5 : 1

5 0

3 years ago