Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Answer:
CH3OH and NADH
Explanation:
The given chemical reaction is an redox reaction in which reduction and oxidation take place.
In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.
In the given redox reaction: CH3OH + NAD --> CH2O + NADH
NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal) by losing electrons.
So, CH3OH (methanol) and NADH are the reduced forms while NAD and CH2O (methanal) are oxidized forms.
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2.34 moles titanium x (6.022 x 10^23)/1 mole titanium = 1.41 x 10^24