<span>1) 0.2M ferric nitrate is added gradually to 1M sodium hydroxide. In result, a red precipitate appears. The precipitate is ferric hydroxide.
2) </span><span>0.2M potassium chromate is added gradually to 0.05M lead acetate. in result, a yellow precipitate appears. The precipitate is called potassium acetate.
The common between the two is that the colors originated from one of the reactants. </span>
Answer:
2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2
Explanation:
First of all, you have to translate the words into an equation.
Fe(iii)2O3 + C ==> Fe + CO2
The easiest way to tackle this is to start with the Oxygens and balance them. They must balance by going to the greatest common factor which is 6. So you multiply the molecule by whatever it takes to get the Oxygens to 6
2 Fe(iii)2O3 + C ==> Fe + 3 CO2
Now work on the irons. There 2 on the left and just 1 on the right. So you need to multiply the iron by 2.
2 Fe(iii)2O3 + C ==> 2 Fe + 3 CO2
Finally it is the turn of the carbons. There are 3 on the right, so you must make the carbon on the left = 3
2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2
And you are done.
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
Answer: 50. 4g
Explanation:
First calculate number of moles of aluminium in 38.8g
Moles = 38.8g/ 26.982mol/g
= 1.44mol
By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.
4 = 2
1.4 = x
Find the value of x
x= (1.4×2)/4= 0.72 mol
0.72 moles of aluminium oxide are produced from 38.8g of aluminium
Now find the mass of aluminium produced.
Mass = moles × molar mass
= 0.72mol × 69.93 mol/g
= 50.4g
Answer:
4.6L
Explanation:
Use the equation (P1*V1)/(T1)=(P2*V2)/(T2)
P= pressure
V= volume
T= temperature in kelvins (remember K= C + 273)
Convert atm to mmHg or vise versa
1.5atm*(760mmhg/1atm)= 1140mmHg
(733mmHg * 5.36L)/(298K)=(1140mmHg * V)/(402K)
V= 4.6 or 4.65L (depending on sig figs)
