Answer:
44.9g
Explanation:
You have to convert grams of CH4 to moles, use the mole-to-mole ratio of CH4 to water, and convert back to grams.
(20.0g CH4)(1 mol CH4/16.04g)(2 mol H2O/1 mol CH4)(18.01 g H2O/ 1 mol) = 44.9127 g
Hope this helps!
The experimental evidence that you could provide that <span> the product isolated is 1-bromobutane would be the changes that happens that are observable by the naked eye. Hope this helps. Have a nice day. Feel free to ask more questions.</span>
Answer:

Explanation:
Let A₀ = the original amount of ⁵⁵Co
.
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.
The general formula for the amount remaining is:
A =A₀(½)ⁿ
where n is the number of half-lives
n = t/t_½
Data:
A = 1.90 ng
t = 45 h
t_½ = 18.0 h
Calculation:
(a) Calculate n
n = 45/18.0 = 2.5
(b) Calculate A
1.90 = A₀ × (½)^2.5
1.90 = A₀ × 0.178
A₀ = 1.90/0.178 = 10.7 ng
The original mass of ⁵⁵Co was
.
Answer:
B.) the number of protons and also the number of electrons in an element