Balanced equation: 2Al+Fe2O3–> Al2O3+2Fe
Using mole calculations you can find that 10g of Al produces 24.3g of Fe under the conditions described
The answer is
option D "CO." Co also known as
Cobalt is the 27th element on the periotic table. It was discovered in <span>1735, it's boiling point is 3200 k.</span>
Atomic mass: 58.9332
Protons: 27
Neutrons: 32
Electrons: 27
Hope this helps!
Answer:
The law is observed in the given equation.
Explanation:
CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂
In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:
- Ca ⇒ 1 on the left, 1 on the right.
- C ⇒ 1 on the left, 1 on the right.
- O ⇒ 3 on the left, 3 on the right.
- H ⇒ 2 on the left, 2 on the right.
- Cl ⇒ 2 on the left, 2 on the right.
As the numbers for all elements involved are the same, the law is observed in the given equation.
Answer:
The exhaust system of the car is the excretory system that removes waste.
The gas in the car is the digestive system that provides the necessary energy.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
