One mole of Fe(NO3)3, or iron(III) nitrate, has three moles of nitrate molecules, which have three moles of oxygen atoms each. We can show this mathematically:
1 mole Fe(NO3)3 * (3 moles NO3)/(1 mole Fe(NO3)3) = 3 moles NO3
3 moles NO3 * (3 moles Oxygen)/(1 mole NO3) = 9 moles Oxygen
9 moles of Oxygen in one mole Fe(NO3)3
The answer is A: When the energy transporting sediments diminishes, the sediments settle in a low-lying area; therefore, deposition always follows erosion
.3 liters... im pretty sure this is correct!!
Answer:
80.27%
Explanation:
Let's consider the following balanced equation.
2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)
First, we have to calculate the moles of Sn²⁺ that react.
We also know the following relations:
- According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
- 1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
- The molar mass of Fe is 55.84 g/mol.
Then, for 1.348 × 10⁻3 moles of Sn²⁺:
If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
