<h3>
<u>moles of H2SO4</u></h3>
Avogadro's number (6.022 × 1023) is defined as the number of atoms, molecules, or "units of anything" that are in a mole of that thing. So to find the number of moles in 3.4 x 1023 molecules of H2SO4, divide by 6.022 × 1023 molecules/mole and you get 0.5646 moles but there are only 2 sig figs in the given so we need to round to 2 sig figs. There are 0.56 moles in 3.4 x 1023 molecules of H2SO4
Note the way this works is to make sure the units are going to give us moles. To check, we do division of the units just like we were dividing two fractions:
(molecules of H2SO4) = (molecules of H2SO4)/1 and so we have 3.4 x 1023/6.022 × 1023 [(molecules of H2SO4)/1]/[(molecules of H2SO4)/(moles of H2SO4)]. Now, invert the denominator and multiply:
<h3 />
Answer:
2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O
Explanation:
Chemical equation:
C₂H₆ + O₂ → CO₂ + H₂O
Balanced chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Step 1:
2C₂H₆ + O₂ → CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 1
H = 12 H = 2
O = 2 O = 3
Step 2:
2C₂H₆ + O₂ → 4CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 2
O = 2 O = 9
Step 3:
2C₂H₆ + O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 2 O = 14
Step 4:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 14 O = 14
0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)
Explanation:
First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume (L)
number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles
We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.
Now to find the mass of hydrated sodium borate we use the following formula:
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g
Learn more about:
molar concentration
brainly.com/question/14106518
#learnwithBrainly
<h3>
Answer:</h3>
81.9 grams
<h3>
Explanation:</h3>
From the question we are given;
- Half-life of C-14 is 5730 years
- Original mass of C-14 (N₀) = 150 grams
- Time taken, t = 5000 years
We are required to determine the mass left after 5000 years
- N = No(1/2)^t/T, where N is the remaining mass, N₀ is the original mass, t is the time taken and T is the half-life.
t/T = 5000 yrs ÷ 5730 yrs
= 0.873
N = 150 g ÷ 0.5^0.873
= 150 g × 0.546
= 81.9 g
Therefore, the mass of C-14 left after 5000 yrs is 81.9 g
