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3 years ago

How many mols in 22.45g of water?

Chemistry

2 answers:

olchik [2.2K]3 years ago

7 0

Answer:

Explanation:

Explanation

Periodic tables vary in how many places of sig digits they show. I always use approximate values which means that you will have to redo the question using your values

Water (1 mole)

2 * H = 1.01 * 2 = 2.02

1 * O = 16 <u>* 1 = 16.00</u>

Mass 1 mole = 18.02

Moles water

given mass = 22.45

molar mass = 18.02

moles = given mass/ molar mass

moles = 22.45 / 18.02

moles = 1.246 moles

Deffense [45]3 years ago

5 0

Answer:

1.247mol

Explanation:

moles=mass/molar mass

molar mass of water = 1+1+16=18

mass = 22.45

22.45÷18

= 1.247 moles of water

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A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio

Pachacha [2.7K]

Answer:

$pK_{a}$ of HA is 6.80

Explanation:

$pK_{a}=-logK_{a}$

Acid dissociation constant ( $K_{a}$ ) of HA is represented as-

$K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

$K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}$

So, $K_{a}=1.6\times 10^{-7}$

Hence $pK_{a}=-log(1.6\times 10^{-7})=6.80$

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3 years ago

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AveGali [126]

The answer is (C) as the definitions states that atomic mass is the number of protons and neutrons.
Hope this helps :).

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3 years ago

Could someone help me with this?

Varvara68 [4.7K]

Solving part-1 only

KMnO_4

Transition metal is Manganese (Mn)

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

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Purple as per the color of potassium permanganate

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2 years ago

Where is each conversion factor found?

Travka [436]

Answer:

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3 years ago

If 0.8675 g of KHP requires 24.56 mL of an NaOH solution to reach the equivalence

ValentinkaMS [17]

Answer:

$M_{base}=0.173M$

Explanation:

Hello there!

In this case, since the titration of acids like KHP with bases like NaOH are performed in a 1:1 mole ratio, it is possible for us to know that their moles are the same at the equivalence point, and the concentration, volume and moles are related as follows:

$n_{acid}=M_{base}V_{base}$

Thus, by solving for the volume of the base as NaOH, we obtain:

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Best regards!

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3 years ago