Answer: This can be quickly solved with "traintracks"
Explanation:
You start w/ grams of water and want to find moles of oxygen gas produced.
So you want to Convert:
Grams of water -> moles of water -> moles of oxygen gas.
The two things you need to know to set up the tracks are:
1)Molar mass of water- H2O
Hydrogen - 1.008(x2)
Oxygen - 16.00
Water - 18.016
Answer : The pressure in torr and in atmospheres are, 745 torr and 0.980 atm respectively.
Explanation :
As we are given that the atmospheric pressure is, 745 mmHg.
Now we have to determine the pressure in torr and atm.
Conversions used:
1 atm = 760 mmHg
1 atm = 760 torr
1 mmHg = 1 torr
As, 760 mmHg = 1 atm
So, 745 mmHg =
and,
As, 1 mmHg = 1 torr
So, 745 mmHg = 745 torr
Therefore, the pressure in torr and in atmospheres are, 745 torr and 0.980 atm respectively.
The following are hazards of corrosion and rancidity;
1. There's reduction of efficiency of the metals.
2. It can lead to contamination especially in the pipes.
3. These can lead to financial losses brought by the destruction of metals.
4. It could result in pollution as a result of chemical reaction bringing about corrosion.
Answer:
The volume is 214, 3 ml
Explanation:
We calculate the weight of 1 mol of NH4OH:
Weight 1 mol NH4OH= Weight N + (Weight H) * 5 + Weight 0
Weight 1 mol NH4OH= 14g + 1 g* 5 + 16g = 35 g/mol
1 mol NH4OH-----35 g
0,6mol NH4OH----X=(0,6mol NH4OHx 35 g)/1 mol NH4OH= 21 g
We have 0,6 mol in 1000ml of solution (0,600 M)
21g NH4OH-----------1000ml
4,50 g NH4OH------x= (4,50 g NH4OH x 1000ml)/21 g NH4OH= 214, 2857143 ml
2HNO₃ + Ba(OH)₂ = Ba(NO₃)₂ + 2H₂O
2 mol HNO₃ - 1 mol Ba(OH)₂
x mol HNO₃ - 2.50×10⁻² mol Ba(OH)₂
x=2.50×10⁻²×2/1=0.05 mol