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Stells [14]
3 years ago
11

What volume of 0.25 M HCl would be needed to neutralize 45 mL of 0.12 M solution of Mg(OH)₂

Chemistry
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

0.38dm³

Explanation:

you just du-mb cuh

I'm jp ig

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Compare amplitudes, wavelengths, and frequencies of waves
mr_godi [17]

Answer:

The answer is option b.

Explanation:

Amplitude is the distance apart each wave is.

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2 years ago
What is the mass number of a single atom of magnesium-25?
Masja [62]

Answer: It is approximately 24.985837

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2 years ago
Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a
Makovka662 [10]

Explanation:

Tollens' reagent is prepared by using two-step process : -

Step 1:

Silver oxide is formed by mixing aqueous silver nitrate with base like sodium hydroxide. The reaction is shown below as:

AgNO_3 + NaOH &\rightarrow AgOH + NHO_3 \nonumber \\ 2AgOH &\rightarrow Ag_2O + H_2O

Step 2

Ammonia solution is drop-wise added until all the silver oxide dissolves to form the reagent. The reaction is shown below as:

Ag_2O + 4NH_3 + H_2O \rightarrow 2Ag(NH_3)_2^+ + 2OH^-

3 0
3 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
3 years ago
Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answ
inessss [21]

Answer:

\Delta S^{0} for the given reaction is -99.4 J/K

Explanation:

Balanced reaction: \frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)

\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]

where S^{0} represents standard entropy.

Plug in all the standard entropy values from available literature in the above equation:

\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K

So, \Delta S^{0} for the given reaction is -99.4 J/K

7 0
3 years ago
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