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2 years ago

The sanlinity of ocean water

Chemistry

1 answer:

Diano4ka-milaya [45]2 years ago

4 0

Answer:

35g/l

Explanation:

Salinity is practically the saltiness of the water, in basic terms. It's basically the amount of salt dissolved in water. Ocean water has a salinity of around 35g/l (that's about 3.5% of which is dissolved salt in water!). However, Atlantic Ocean (the ocean with the saltiest water), can range up to 37g/l.

<em>Feel free to mark this as brainliest :D</em>

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Given the unbalanced equation: ___Al2(SO4)3+___Ca(OH)2—>___Al(OH)3+__CaSO4

nasty-shy [4]

Answer:

The answer to your question is letter B. 9

Explanation:

Unbalanced reaction

Al₂(SO₄)₃ + Ca(OH)₂ ⇒ Al(OH)₃ + CaSO₄

Reactants Elements Products

2 Al 1

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Al₂(SO₄)₃ + 3Ca(OH)₂ ⇒ 2Al(OH)₃ + 3CaSO₄

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2 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

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The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

2 years ago

The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl

AlekseyPX

Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

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The given balanced equilibrium reaction is,

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

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At eqm. conc. (x-2y) M (y) M (3y) M

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Now put all the given values in this expression, we get :

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

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$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M

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3 years ago