1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
A conversion factor is a fraction or a ratio representing a relationship of two different measurement values. To write 20% m/v to a conversion factor, we need to remember that a percent is a value that represents the amount of a part per 100 units of the whole. M/v in the given value represents that the percentage is by mass per volume. So, to write it as a conversion factor, we do as follows:
20% m/v = 20 mass units / 100 volume units = 1 mass units / 5 volume units
Usually units of this are in g per L. So, it is equivalent to 1 g / 5 L
Answer:
53.11× 10²³ molecules
Explanation:
Given data:
Number of molecules of CO₂ = ?
Mass of CO₂ = 388.1 g
Solution:
Formula:
Number of moles = mass/ molar mass
Molar mass of CO₂ = 12× 1 + 16×2
Molar mass of CO₂ = 44 g/mol
Now we will put the values in formula.
Number of moles = 388.1 g/ 44 g/mol
Number of moles = 8.82 moles
Now we will calculate the number of molecules by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
8.82 mol × 6.022 × 10²³ molecules / 1 mol
53.11× 10²³ molecules
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j