Answer:
2 PO₄³⁻(aq) + 3 Fe²⁺(aq) ⇒ Fe₃(PO₄)₂
Explanation:
Let's consider the complete ionic equation between the ions present. It includes all the ions and the insoluble compounds (Fe₃(PO₄)₂ is insoluble).
Na⁺(aq) + 2 PO₄³⁻(aq) + 3 Fe²⁺(aq) + NO₃⁻(aq) ⇒ Fe₃(PO₄)₂ + Na⁺(aq) + NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble compounds.
2 PO₄³⁻(aq) + 3 Fe²⁺(aq) ⇒ Fe₃(PO₄)₂
Answer:
![[Fe^{+3}]=0.700 M](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D0.700%20M)
![[NO_{3}^{-}]=2.10 M](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D2.10%20M)
Explanation:
Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:
Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.
Solving for [Fe],
![[Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D%5Cfrac%7BC_%7B1%7D.V_%7B1%7D%7D%7BV_%7B2%7D%20%7D)
![[Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D%5Cfrac%7B%281.75%29.%2830.0%29%7D%7B45.0%2B30.0%20%7D)
For [NO₃⁻],
There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.
![[NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D%5Cfrac%7BC_%7B1%7D.V_%7B1%7D%7D%7BV_%7B2%7D%20%7D)
![[NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D%5Cfrac%7B%285.25%29%2830.0%29%7D%7B30.0%2B45.0%20%7D)
Answer:
Total pressure = 16.42× 10⁻⁹atm
Explanation:
Given data:
Moles of H₂ = 2.50 × 10⁻³ mol
Moles of He = 1.00 × 10⁻³ mol
Mass of Ne = 3 × 10⁻⁴ mol
Volume = 10 L
Temperature = 35°C
Total pressure = ?
Solution:
Pressure of hydrogen:
P = nRT / V
P = 2.50 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 63.22× 10⁻³ atm. L /10 L
P = 6.3 × 10⁻³atm
Pressure of helium:
P = nRT / V
P = 1.00 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 25.29 × 10⁻³ atm. L /10 L
P = 2.53× 10⁻³ atm
Pressure of neon:
P = nRT / V
P = 3 × 10⁻⁴ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L
p = 75.86× 10⁻³ atm. L /10 L
P = 7.59× 10⁻³ atm
Total pressure of mixture:
P(mixture) = pressure of hydrogen + pressure of helium+ pressure of neon
P(mixture) = 6.3 × 10⁻³atm + 2.53× 10⁻³ atm + 7.59× 10⁻³ atm
P(mixture) = 16.42× 10⁻⁹atm
Answer:
Explanation:
mass of one virus = 9.0 x 10⁻¹² mg
mass of one mole = 6.02 x 10²³ x mass of one virus
= 6.02 x 10²³ x 9.0 x 10⁻¹²
= 54.18 x 10¹¹ mg
= 54 x 10⁸ g .
= 54 x 10⁵ kg .
b )
let n be no of moles of virus that will be equal to weight of oil tanker
n x 54 x 10⁵ = 3 x 10⁷
n = 5.5555
rounding off to 2 significant figure
5.6 moles Ans .
Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
