I think B
Hope this helps!
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
The correct answer is b, 18 protons
1.39 g HCl
Explanation:
The balanced chemical equation for this reaction is given by
Zn(<em>s</em>) + 2HCl(<em>aq</em>) ---> ZnCl2(<em>aq</em>) + H2(<em>g</em>)
Convert the # of grams of Zn to moles:
1.25 g Zn × (1 mol Zn/65.38 g Zn) = 0.0191 mol Zn
Use the molar ratio to find the # of moles of HCl needed to react completely with the given amount of Zn:
0.0191 mol Zn × (2 mol HCl/1 mol Zn) = 0.0382 mol HCl
Convert this amount to grams:
0.0382 mol HCl × (36.458 g HCl/1 mol HCl) = 1.39 g HCl
Answer:
Average atomic mass = 85.557 amu.
Explanation:
Given data:
Percent abundance of Rb-85 = 72.15%
Percent abundance of Rb-87 = 27.85%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (72.15×85)+(27.85×87) /100
Average atomic mass = 6132.75 + 2422.95 / 100
Average atomic mass = 8555.7 / 100
Average atomic mass = 85.557 amu.
