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Evgen [1.6K]
3 years ago
7

What is the concentration of lithium ions in 0.400 M Li₂HPO₄?

Chemistry
1 answer:
omeli [17]3 years ago
3 0

Answer: 0.800 M

Basically, the ratio is 2:1, so when you have 1 M of Li2HPO4 it means you have 2 M in the concentration of lithium ions. So knowing the ratio, we know we just need to multiply 0.400 x 2.

I'm bad at explaining it, but got it right on chem101 sooo

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

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Answer:

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Explanation:

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At pH 2.5, its activity will decrease enormously as it becomes denatured by the high acidity of the stomach contents.

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Which statement is most consistent with the plum pudding model of the atom?
lianna [129]
The answer to this question would be: <span>1) Electrons occupy regions of space
</span><span>
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Answer:

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