The answer for the question given is A. Heating with a flame.
One of the processes that cause a chemical change is heating. An example of this is cooking or baking, which causes a permanent chemical change in substances. Also, all of the other choices are examples of physical changes.
Answer:
Volume of ammonia 3 L.
Explanation:
Given data:
Temperature = 25°C ( 25+273= 298 k)
Pressure = 99.5 kpa (99.5/101 = 0.98 atm)
Volume of nitrogen = 1.50 L
Volume of ammonia = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Moles of nitrogen:
PV = nRT
n = PV/RT
n = 0.98 atm × 1.50 L / 0.0821 atm. L/mol. K × 298 K
n = 1.47 /24.5 /mol
n = 0.06 mol
Now we will compare the moles of nitrogen with ammonia.
N₂ : NH₃
1 : 2
0.06 : 2×0.06 = 0.12 mol
Volume of ammonia:
PV = nRT
V = nRT/P
V = 0.12 mol× 0.0821 atm. L/mol. K × 298 K/ 0.98 atm
V = 2.9 atm. L /0.98 atm
V = 3 L
Answer:
mass of hydrogen collected is 0.016 gram
Explanation:
Given values:
For calculating mass we have to find the number of moles first
Ideal gas equation PV =nRT
Volume= 195 ml
Pressure: 753 torr =0.99 atm
Temperature: 25+273= 298 Kelvin
Ideal gas constant R= 0.0821 Latm/molK
Number of moles n= ?
So n= PV/RT
Adding the values
n= = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(273 K)
n = 0.007897 moles of H₂
Now mass of hydrogen collected = number of moles x Molar mass of H₂
= 0.007897 x 2
= 0.0157 g H₂
Now mass of hydrogen collected is 0.016 gram (rounding the amount)
Answer:
a) 0.13 bar
b) 5.05 L
Explanation:
Let us take the mass of the mixture to be 100 g. Hence, 70% by weight of dioxygen corresponds to 70 g
Mass of dihydrogen = 100g - 70 g = 30g
Number of moles of dioxygen = 70g/32 g/mol = 2.2 moles of dioxygen
Number of moles of dihydrogen = 30g/2g/mol = 15 moles of dihydrogen
Total number of moles = 2.2 + 15 = 17.2 moles
Mole fraction of dioxygen = 2.2/17.2 = 0.13
Partial pressure = mole fraction * total pressure
Partial pressure of dioxygen = 0.13 * 1 = 0.13 bar
ii) number of moles in 8.8 g of CO2 = 8.8g/44g/mol = 0.2 moles
T = 31.1 + 273 = 304.1 K
P = 1 bar
V= ?
R = 0.083 bar L K-1 mol-1
From
PV=nRT
V = nRT/P
V= 0.2 * 0.083 * 304.1/1
V= 5.05 L
The total number of electron pairs that are shared between the two carbon atoms in a molecule of ethyne is 3.