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gtnhenbr [62]
2 years ago
4

Which answer choice represents a balanced alpha emission nuclear equation? Superscript 52 subscript 26 upper F e right arrow sup

erscript 52 subscript 25 upper M n plus superscript o subscript plus 1 e. Superscript 235 subscript 92 upper U right arrow superscript 239 subscript 94 upper P u s plus superscript 4 subscript 2 upper H e. Superscript 189 subscript 83 upper Bi right arrow superscript 185 subscript 81 upper T l plus superscript 4 subscript 2 H e. Superscript 14 subscript 6 upper C right arrow superscript 14 subscript 7 upper N plus superscript 0 subscript negative 1 e.
Chemistry
2 answers:
lawyer [7]2 years ago
7 0

Answer:

Superscript 189 subscript 83 upper Bi right arrow superscript 185 subscript 81 upper T l plus superscript 4 subscript 2 H e.

Explanation:

aev [14]2 years ago
5 0

Answer:

d.

Explanation:

28 .     Ra

88

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After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ___
mamaluj [8]

Answer:

The answer to be filled in the respective blanks in question is

3 and 1

Explanation:

So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.

Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1

 

7 0
2 years ago
"The elementary reaction 2 NO2(g) → 2 NO(g) + O2(g) is second order in NO2 and the rate constant at 600 K is 6.77 × 10-1 M-1s-1.
blsea [12.9K]

Answer : The half-life at this temperature is, 3.28 s

Explanation :

To calculate the half-life for second order the expression will be:

t_{1/2}=\frac{1}{k\times [A_o]}

When,

t_{1/2} = half-life = ?

[A_o] = initial concentration = 0.45 M

k = rate constant = 6.77\times 10^{-1}M^{-1}s^{-1}

Now put all the given values in the above formula, we get:

t_{1/2}=\frac{1}{6.77\times 10^{-1}M^{-1}s^{-1}\times 0.45M}

t_{1/2}=3.28s

Therefore, the half-life at this temperature is, 3.28 s

7 0
2 years ago
Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?
Margarita [4]

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

6 0
2 years ago
Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o
Marizza181 [45]

Answer:

30.6 g of C is formed.

Explanation:

2A + B → C

Average rate of reaction = 2[A]/Δt = [B]/Δt = [C]/Δt

Average rate of reaction = [C]/Δt

Average rate of reaction  = 15 g / 9 min

Average rate of reaction  = 1.7 g of C / min

Average rate of reaction = [C]/Δt

[C] = Average rate of reaction x Δt

[C] = 1.7 g of C / min x 18 min

[C] = 30.6 g of C

4 0
2 years ago
When dissolved beryllium chloride reacts with dissolved silver nitrate in water, aqueous beryllium nitrate and silver
lisabon 2012 [21]

Answer:

The balanced chemical equation is :

BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl

Explanation:

The chemical equation in which number of atoms in reactants is equal to products is called balanced equation.

Formula of , Beryllium Chloride :[BeCl_{2}

Beryllium Nitrate : BeNO_{3}

Silver Chloride : AgCl

Silver Nitrate  :AgNO_{3}

When beryllium chloride reacts with dissolved silver nitrate in water , following reaction occur :

 BeCl_{2} + 2AgNO_{3}\rightarrow Be(NO_{3})_{2} + 2AgCl

The number of atoms in reactant as well as in products are balanced :

Be = 1

Ag = 2

N =2

O = 6

Cl = 2

4 0
2 years ago
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