<span>Balanced nuclear reaction: 234/91 Pa → 4/2 He + 230/89 Ac.
1) number of protons and neutrons on both side of nuclear reaction must be the same. There are 91 protons (atomic number) and 143 neutrons (mass number - atomic number, 234 - 91 = 143 or (4-2) + (230 - 89) = 143)) on both side of reaction.
2) a</span>lpha
decay is radioactive decay<span> in which
an atomic nucleus emits
an alpha particle (helium
nucleus) and transforms into an atom with an atomic
number that is reduced by two and mass number that is reduced by four, so atomic mass of new element is 89 (91 - 2) and mass number is 230 (234 - 4 = 230).
3) look at atomic number, element with atomic number 89 is actinium.</span>
Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives 
= 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = 
= 
= 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
