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ollegr [7]
3 years ago
9

Which of the following is true for a nuclear reaction? Electrons are lost Electrons are gained Identity of element changes Ident

ity of element remains same
Chemistry
2 answers:
yawa3891 [41]3 years ago
8 0

Answer: Identity of element changes

Explanation:

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Transmutation is a chemical reaction in which nucleus of the atom undergoes a change by radioactive decay and thus result into formation of new nucleus. Thus the identity of the element changes.

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{56}^{143}Ba+_{36}^{90}Kr+3_0^1\textrm{n}

anyanavicka [17]3 years ago
5 0

All nuclear chemistry revolves around changing the identity of base elements.

The answer is the third statement.

Because nuclear chemistry is changing base elements' identities, the atomic number/number of protons is what is changing.  While the number of electrons can change, your question is too general to know whether electrons are going to be gained, lost, or retained in the reaction.

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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.

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First we need to calculate moles of each

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

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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

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pH = 7.21 + log 0.225 / 0.162

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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
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<h3> Further explanation </h3>

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Hi, there the answer is

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