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3 years ago

G- how many grams of f2 gas are there in a 5.00-l cylinder at 4.00 × 103 mm hg and 23°c?

1 answer:

6 0

PV = nRT (4000 Torr)(5 L) = n (62.4 Torr-L/mol-K)(296K) n = 1.08 moles 28 g/mol, 1.08 moles = 30.3 grams your answer is C.30.3 g

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If a given aqueous solution at 298 K contains [H+] = 1.0 × 10-9, what is the [OH-]?

Zinaida [17]

Answer:

Explanation:

$\frac{1x10}x^{-14} = 1x10^{-9} \\ x =1x10^{-5} \\\\[OH][H]= 1x10^{-14}$

The concetration can be found by dividing the water ph constant by the [H=] or [OH] to find the other

6 0

3 years ago

What is the pH of a 0.028M solution (pH= -log(M)) A) 3.56 B) 2.88 C) 1.55 D) 1

kupik [55]

Answer:

1.55

Explanation:

-log(M)=pH

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3 years ago

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Brilliant_brown [7]

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2 years ago

The equilibrium constant for the reaction

FinnZ [79.3K]

Answer: The concentrations of $Cl_2$ at equilibrium is 0.023 M

Explanation:

Moles of $Cl_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol$

Volume of solution = 1 L

Initial concentration of $Cl_2$ = $\frac{0.14mol}{1L}=0.14M$

The given balanced equilibrium reaction is,

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial conc. 0.14 M 0 M 0M

At eqm. conc. (0.14-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}$

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

$4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}$

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of $Cl_2$ at equilibrium is 0.023 M

7 0

3 years ago

A chemist has dissolved a certain substance in water. The chemist knows that more of the substance could be dissolved into the w

FrozenT [24]

Unsaturated………. The answer

5 0

2 years ago

Read 2 more answers