Answer:
1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.
2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.
The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.
This are some of the transuranium elements:
Neptunio - 93
Plutonium - 94
Americium - 95
Curium - 96
Berkelium - 97
Californium - 98
Einstenium - 99
And so all the known elements (the last one is the 118).
3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.
4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:
0
β, which means 0 mass and charge - 1.
-1
5) This is, then, an example of a β decay equation for one transuranium element:
239 239 0
Np → Pu + β
93 94 -1
As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.
Explanation:
Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Is a compound that went through a chemicAl change
Do you mean h=7.0+10? If so your answer is 70.
