Two ways that salinity can increase

Chemistry

1 answer:

german2 years ago

3 0

Answer:

Evaporation of ocean water and the formation of sea ice both increase the salinity of the ocean.

Explanation:

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Need help with 22 and 24

choli [55]

Answer:

22:

Formular:

$atomic \: mass = \frac{ \sum(isotopic \: mass \times \%abundance)}{100} \\$

substitute:

$atomic \: mass = \frac{(23.985 \times 78.70) + (24.986 \times 10.13) + (25.983 \times 11.17)}{100} \\ \\ = \frac{(1887.620) + (253.108) + (290.230)}{100} \\ \\ = \frac{2430.958}{100} \\ \\ { \boxed{ \boxed{average \: atomic \: mass = 24.3 \: amu}}}$

23:

Same element is represented by same number of protons. 

Answer:

6 protons. 6 protons

7 neutrons. 8 neutrons

6 electrons. 6 electrons

Note: Atoms with same proton number but different mass number are called isotopes

5 0

2 years ago

Read 2 more answers

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

Answer:

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$