The increase of the boling point of a solution is a colligative property.
The formula for the increase of the normal boiling point of water is:
ΔTb = Kb * m
Where m is the molallity of the solution and Kb is the molal boiling constant in °C/mol.
ΔTb = 0.51 °C / m * 0.100 m = 0.051 °C.
So, the new boiling temperature is Tb = 100°C + 0.051°C = 100.051 °C.
Answer: 100.051 °C
Answer:
Moles of H₂S needed = 6.2 mol
Moles of SO₂ produced = 6.2 mol
Explanation:
Given data:
Number of moles of O₂ = 9.3 mol
Moles of H₂S needed = ?
Moles of SO₂ produced = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Now we will compare the moles of oxygen with H₂S.
O₂ : H₂S
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
Now we will compare the moles of SO₂ with both reactant.
O₂ : SO₂
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
H₂S : SO₂
2 : 2
6.2 : 6.2 mol
So 6.2 moles of SO₂ are produced.
Because it’s kinetic energy INCREASES the speed
Answer:
<u>ATGGCCTA</u>
Explanation:
For this we have to keep in mind that we have a <u>specific relationship between the nitrogen bases</u>:
-) <u>When we have a T (thymine) we will have a bond with A (adenine) and viceversa</u>.
-) <u>When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa</u>.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
<u>ATGGCCTA</u>
<u></u>
I hope it helps!
