Answer:
79.8g/dm³
Explanation:
As you can see, the solution in the problem contains 0.5 moles of copper sulfate per dm³. To solve this question we must convert these moles to grams using its molar mass (Molar mass CuSO4 = 159.609g/mol) as follows:
0.5mol CuSO4/dm³ * (159.609g/mol) =
<h3>79.8g/dm³</h3>
Thermal (heat) energy is ur answer mate
Hope it helps
The moles of Ba(OH)2 that is required to react with 117 HBr is calculated as below
find the moles of HBr used
mass/ molar mass = 117 g/ 80.9 g/mol = 1.446 moles
write the reacting equation
Ba(OH)2 + 2 HBr = BaBr2 + 2 H2O
by use of mole ratio of Ba(OH)2 : HBr which is 1:2 the moles of Ba(OH)2 is therefore
= 1.446 moles x1/2 = 0.723 moles of Ba(OH)2
Answer:
the channel tunnel has been a very busy and challenging experience and we are mot doing.
Explanation:
ok and I like the country studies opportunity and I will be found by him too as he is a fact in the sentence.
Answer:
The atomic weight of hypothetical element will be 63.568 amu.
Explanation:
Given data:
First isotope mass = 62.2 amu
Percentage abundance of first isotope = 24%
Mass of second isotope = 64 amu
Percentage abundance of second isotope = 100- 24 = 76%
Solution:
The atomic weight of hypothetical element will be the average atomic mass of its isotopes.
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
Now we will put the values in formula.
Average atomic mass = [24 ×62.2] + [76× 64] / 100
Average atomic mass = 1492.8 + 4864 / 100
Average atomic mass = 6356.8/100
Average atomic mass = 63.568 amu