Neptunium's only naturally occurring isotope, 23793np, decays by emitting one alpha particle, one beta particle, and one gamma ray. What is the new atom formed from this decay?
Answers
The given isotope of Neptunium is
Alpha decay of an isotope results in daughter nuclide with mass number less by 4 units and atomic number less by 2 units than the parent isotope.
Alpha decay of Neptunium-237 can be represented as:
Beta decay of the above formed protactinium nuclide can be represented as:
Gamma decay releases only energy in the form of gamma rays, the nuclide remains the same.
γ
The new atom formed after the given decays is
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What is the total number of moles of C that must completely react to produce 2.0 moles of C2H6?
(1) 1.0 mol (3) 3.0 mol
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Answers
Answer: The correct answer is option 4.
Explanation:
According to the reaction, 1 mol of is obtained from 2 moles of C gives
Then, 2 moles of will be obtained from:
moles of C =4 moles
Hence, the correct answer is option 4.
Answers
# moles = mass / molar mass
Molar mass calculation
Barium hydroxide = Ba (OH)2
Atomic masses
Ba = 137.4 g/mol
O=16 g/mol
H=1 g/mol
Molar mass of Ba (OH)2 = 137.4 g/mol + 2*16g/mol + 2*1 g/mol = 171.4 g/mol
# mol = 25.0g/171.4 g/mol = 0.146 mol
For the volume of water use the fact that the density is 1g/ml., so 120 g = 120 ml = 0,120 liters.
M = 0.146mol / 0.120 liters = 1.22 mol/liter
Answers
cold water with no salt
B.
warm water with no salt
C.
warm, salty water
D.
cold, salty water
Answers
Answers
Answer: Option (d) is the correct answer.
Explanation:
It is known that sea water contains a large amount of common salt, that is, sodium chloride (NaCl). This salt is basic in nature.
Since, it is also known that species which have pH equals to 7 are basic in nature. Species which have pH less than 7 are acidic in nature and species which have pH greater than 7 are basic in nature.
Hence, due to the presence of common salt in sea water its surface is basic in nature.
Therefore, we can conclude that the normal pH of surface sea water is 9.0.
Hope this helps
Answers
Answer:
Second order
Δ[ClO⁻]/Δt = - 4.183 x 10⁻⁴ M/min
Explanation:
Given the data:
Experiment # [ClO–] (M) Initial Rate of Formation of ClO3– (M/min)
1 10.452 1.048 x 10⁻⁴
2 20.903 4.183 x 10⁻⁴
we need to determine the order of the reaction with respect to ClO⁻.
We know the rate law for this reaction will have the form:
Rate = k [ClO⁻]^n
where n is the order of the reaction. Thus, what we need to do is to study the dependence of the initial rate on n for the experiment.
If the reaction were zeroth order the rate would not change, so we can eliminate n= 0
If the reaction were first order, doubling the concentration of [ClO–] , as it was done exactly in experiment # 2, the initial rate should have doubled, which is not the case.
If the reaction were second order n: 2, doubling the concentration of [ClO–] , should quadruple the initial rate of formation of ClO3–, which is what it is observed experimentally. Therefore the reaction is second order respect to ClO–.
The initial rate of consumption of ClO⁻ is the same as the rate of formation of ClO₃⁻ since:
Δ = - Δ[ClO⁻]/Δt = + Δ[ClO₃⁻]/Δt = + 1/2 [Cl⁻] /Δt
where t is the time.
from the coefficients of the balanced chemical equation.
- Δ[ClO⁻]/Δt = + Δ[ClO₃⁻]/Δt = + 1/2 [Cl⁻ ] = rate
Δ[ClO⁻]/Δt = - 4.183 x 10⁻⁴ M/min