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AlekseyPX
3 years ago
14

Transuranium elements are artificial and may be prepared by a fusion process in which heavy nuclei are bombarded with light ones

. Complete the equation for the following fusion reaction by filling in the nuclear symbol for the missing product. 343Am90 + 4He2 → 2n1/0
Chemistry
1 answer:
Scilla [17]3 years ago
7 0

Answer:

_{97}^{345}\textrm{Bk}

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

Am has an atomic number of 95. So correct reaction is:-

^{343}_{95}\textrm{Am}+^{4}_{2}\textrm{He}\rightarrow ^A_Z\textrm{X}+2^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

343 + 4 = A + 2

A = 345

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

95 + 2 = Z + 0

Z = 97

Hence, the isotopic symbol of unknown element is _{97}^{345}\textrm{Bk}

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There will be more light-colored wolves.

Explanation:

Evolution is based off the survival of those with the most advantages in the wild. Since their light coat gives them a survival advantage, they will reproduce more.

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2 years ago
a molecule in a chemical reaction that has its chemical bonds broken is called the _______ in the reaction?​
nika2105 [10]

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reactants

Explanation:

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2 years ago
Hydrazine (N2H4) is a fuel used by some spacecraft. It is normally oxidized by N2O4 according to the following equation: N2H4(l)
vitfil [10]

Answer:

The enthalpy of the reaction is coming out to be -380.16 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

N_2H_4(l)+N_2O_4(g)\rightarrow 2N_2O(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]

We are given:

\Delta H_f_{(N_2O)}=81.6 kJ/mol\\\Delta H_f_{(H_2O)}=-241.8 kJ/mol\\\Delta H_f_{(N_2H_4)}= 50.6 kJ/mol\\\Delta H_f_{(N_2O_4)}=9.16 kJ/mo

Putting values in above equation, we get:

\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ

Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.

6 0
3 years ago
Why do you think that the electrons are not shown?
snow_lady [41]
Because there to tiny to see
7 0
3 years ago
Question 21
Mumz [18]
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.

Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
5 0
2 years ago
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