Answer:
V = 44.85 L
Explanation:
Given data:
Volume of H₂ = ?
Number of moles of H₂ = 2.0 mol
Given temperature = 273.15 K
Given pressure = 1 atm
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 2.0 mol × 0.0821 atm.L/ mol.K × 273.15 K
V = 44.85 atm.L / 1 atm
V = 44.85 L
C6H12O6(s) + 9 O2(g)
-> 6 CO2(g) + 6 H2O(g)
There are 9 moles of O2 needed for the complete combustion of glucose.
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
Answer:
24.7 grams H₂O
Explanation:
To find the mass of water, you should (1) convert from grams HCl to moles (via molar mass from periodic table), then (2) convert from moles HCl to moles H₂O (via mole-to-mole ratio from equation), and then (3) convert from moles H₂O to grams (via molar mass from periodic table).
1 HCl + KOH --> 1 H₂O + KCl
Molar Mass (HCl): 1.008 g/mol + 35.45 g/mol
Molar Mass (HCl): 36.458 g/mol
Molar Mass (H₂O): 2(1.008 g/mol) + 16.00 g/mol
Molar Mass (H₂O): 18.016 g/mol
50 g HCl 1 mole HCl 1 mole H₂O 18.016 g
-------------- x ------------------ x ------------------- x ------------------- = 24.7 g H₂O
36.458 g 1 mole HCl 1 mole H₂O
Distilled water.high school Lab
