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VMariaS [17]
3 years ago
13

A radioactive isotope of vanadium, V, decays by producing a  particle and gamma ray. The nuclide formed has the atomic number:

A) 22 B) 21 C) 23 D) 24 E) none of these
Chemistry
1 answer:
Slav-nsk [51]3 years ago
6 0

C) 24

Explanation:

When a radioactive isotope of Vanadium, V decays by producing a beta particle and gamma rays, the atomic number will be 24.

 Vanadium is the 23rd element on the periodic table.

It has a mass number of 51.

              atomic number is 23

                                      ⁵¹₂₃V

  A beta particle is symbolized as °₋₁B

   Gamma rays are represented by Y

Now to the equation:

         ⁵¹₂₃V    →    ₐⁿK   +   °₋₁B  +    Y

In a nuclear reaction the mass an atomic number is conserved

   51 = n + 0

            n = 51

   23 = a -1

        a = 24

We can see that the nuclide formed will have an atomic number of 24

learn more:

Transmutation brainly.com/question/3433940

#learnwithBrainly

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If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain
skad [1K]

Answer:

11.31 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>

<em></em>

<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>

6 0
3 years ago
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Some glucose produced by gluconeogenesis is stored in the body as glycogen. Order the steps of glycogen synthesis.
Harlamova29_29 [7]

Answer:

e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi

a. Pyrophosphatase converts PPi and water into two Pi

b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP

Explanation:

Glycogen synthesis or glycogenesis is the process of synthesis of glycogen molecules from glucose molecules in living organisms. Glycogen is a polysaccharide storage form of glucose and helps to store excess glucose in the body form use when required by the body.

The synthesis of glycogen involves sugar nucleotides. Sugar nucleotides are compounds in which a sugar molecule is attached to a nucleotide through phosphate ester bond, resulting in the activation of the sugar molecule. The sugar nucleotides then are used as substrates for the polymerization of the monosaccharide sugars into disaccharides, oligosaccharides and polysaccharides.

In the synthesis of glycogen, glucose-6-phosphate from phosphorylation of free glucose by hexokinase is first isomerized to glucose-1-phosphate by phosphoglucomutase.

Glucose-1-phosphate is then converted to UDP-glucose by its reaction with UTP catalyse by UDP-glucose pyrophosphorylase. The reaction is favoured by the rapid hydrolysis of PPi produced to two molecules of inorganic phosphate by the enzyme pyrophosphatase.

Glycogen synthase then adds a glucose unit from UDP-glucose to a growing chain of glycogen, producing a larger glycogen molecule and free UDP.

6 0
3 years ago
Determine how many formula units are in 2.35 moles of lithium phosphite?
Stells [14]

Answer:

Determine how many formula units are in 2.35 moles of lithium phosphite?

Explanation:

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4 0
3 years ago
For an endothermic reaction, how will the value for Keq change when the temperature is increased?
Nataly [62]
I put the answer <em>C: Keq will increase</em>, on PLATO. Hope this works for you!
4 0
3 years ago
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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
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