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allsm [11]
3 years ago
9

Give two examples of activities performed by the veterinarian

Chemistry
1 answer:
AURORKA [14]3 years ago
8 0

Answer:

exp:

two ways of activities performed by a veterinarian is:

performing surgery and checkups.

You might be interested in
Examples of a pure substances
DiKsa [7]

Answer:

Explanation:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.

3 0
2 years ago
Read 2 more answers
HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i
OleMash [197]

<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

<u>Explanation:</u>

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

HNO_3+S\rightarrow H_2SO_4+NO

<u>On the reactant side:</u>

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

<u>On the product side:</u>

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

3 0
3 years ago
1. Given the specific heat of lead is 0.129 J/g.C and that it takes 93.4J of energy to
zheka24 [161]

Answer: 40 grams

Explanation:

The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = 93.4J

M = ?

C = 0.129 J/g.C

Φ = 40.4°C - 22.3°C = 18.1°C

Then, Q = MCΦ

Make Mass, M the subject formula

M = Q/CΦ

M = (93.4J) / (0.129 J/g.C x 18.1°C)

M = 93.4J / 2.33J/g

M = 40 g

Thus, the mass of the lead is 40 grams

8 0
3 years ago
A nuclide of phosphorus 3015P decays into a nuclide of silicon 3014Si. In order to satisfy charge conservation and lepton number
jeka94

In order to satisfy charge conservation and lepton number conservation the other products must be neutron.

<h3>What is conservation of mass?</h3>

The principle of conservation of mass states that, the sum of the initial mass of reactants must be equal to final mass of the products.

_{15}^{30}P\ --- > \ _{14}^{30} Si \ + \ ^{0}_{1} n

The balanced reaction of radioactive decay of phosphorous shows conservation of mass.

Thus, in order to satisfy charge conservation and lepton number conservation the other products must be neutron.

Learn more about radioactive decay here: brainly.com/question/1383030

#SPJ1

3 0
2 years ago
Do cells divide twice in asexual reproduction, sexual reproduction, or both?
igor_vitrenko [27]
Cells divide in asexual reproduction!


I hope this helps 



please mark me as brainliestt!!
8 0
3 years ago
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