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Genrish500 [490]

3 years ago

12

What is the difference between oxygen-16 and oxygen-18

1 answer:

Ulleksa [173]3 years ago

8 0

Answer:

Oxygen 16 has 8 protons and 8 neutrons while Oxygen 18 has 8 protons and 10 neutrons . These are isotopes of oxygen i.e same no. of protons but different no. of neutron .

Hope You Understood .

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Which state of matter has a definite volume, but does not have a definite shape?

evablogger [386]

Answer:

Liquid, definite volume but no definite shape.

Explanation:

7 0

3 years ago

The element phosporus is located to the right of the element magnesium along the same period in the periodic table.

Gala2k [10]

I’m assuming this is a true and false question. The answer is true. Phosphorus has a larger atomic mass. Phosphorus is number 15 on the periodic table. Phosphorus is a non-metal.

7 0

3 years ago

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Westkost [7]

Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

c)

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

d)

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

3 years ago

Pls help . I don't know this one .

aliya0001 [1]

The answer is the first one, Xe

5 0

4 years ago

Will give brainliest!!!!

Alekssandra [29.7K]

Answer:

Its B :)

Hope this helps

7 0

3 years ago