Answer:
What type of work do u have? LOL
Explanation:
<span>First we can calculate the area of the rectangular lawn using the formula:
Area = Width x Length = 21 ft x 20 ft = 420 square feet
And the total number of snow flakes per minute on the entire lawn is:
(1350 snowflakes per minute per square foot) x (420 square feet) = 567,000 snowflakes per minute
In one hour (or 60 minutes) we get a total of:
(567,000 snowflakes per minute) x (60 minutes / 1 hour) = 34,020,000 snowflakes
The total mass of which would be:
34,020,000 snowflakes x 1.60 mg = 54,432,000 mg = 54.432 kg (as 1 kg = 1,000,000 mg).
So 54.432 kg of snow accumulates every hour on the lawn.</span>
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
.</span>
27.6 - 22.3 = 5.3; you simply subtract the old density from the new density.
M(P)=3.72 g
M(P)=31 g/mol
m(Cl)=21.28 g
M(Cl)=35.5 g/mol
n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol
n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol
P : Cl = 0.12 : 0.60 = 1 : 5
PCl₅ - is the empirical formula