Answer:
25.0 mol O₂ are required in the second reaction
Explanation:
N₂ (g) + 3H₂ (g) → 2NH₃ (g)
4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6 H₂O(l)
Molar ratio in first reaction is 1:2
For every mol of N₂. I make 2 moles of ammonia. If I have 20 moles of N₂, i'm going to get, 40 moles of ammonia.
In the second reaction, molar ratio between products is 4:5.
If I obtained 40 moles of ammonia in first step, let's prepare the rule of three.
4 moles of ammonia react with 5 moles of O₂
40 moles of ammonia react with ( 40.5) /4 = 25moles
Answer:
Explanation:
Using the necessary reagents to faciliate the synthesis of the organic compounds as shown in the attached file.
It is
A Venus flytrap closes on a fly.
C. A plant grows toward the sunlight.
E.Bright lights cause a baby to blink.
( i had a test have a good day )
Answer:
The Kinetic Energy is approximately 3 times decreased
Explanation:
A baseball weighs 5.13 oz.
a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?
b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.
Kinetic Energy (KE)=0.5×mass×velocity ^ 2
Kinetic Energy (KE)=0.5×mass × velocity ^ 2
Joules = kg×m^2/s^2
1 mile = 1609.344 meters
1 hour = 3600 sec
1 Oz = 28.34952 g = 0.02834952 kg
a) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=130.761 kg×m^2/s^2 = 130.761 Joules
b) KE=0.5×m×v^2
=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2
=43.51028 kg×m^2/s^2 = 43.51028 Joules
= 130.761 / 43.51028 = 3.00528,
As such the Kinetic Energy is approximately 3 times decreased
Answer:
Mass% Cr = 85.5%
Explanation:
<u>Given:</u>
Mass of CrBr3 sample = 0.8409 g
Mass of the AgBr precipitate = 1.0638 g
<u>To determine:</u>
The mass percent of Cr in the sample
<u>Calculation:</u>
The reaction of CrBr3 with silver nitrate results in the precipitation of the bromide ion as silver chloride (AgBr) and Cr as soluble Cr(NO3)2
CrBr3(aq) + 3AgNO3(aq)→ 3AgBr(s) + Cr(NO3)3(aq)
Molecular weight of AgBr =187.77 g/mol
Moles of AgBr precipitated is:
Since 1 mole of AgBr contains 1 mole of Cl, therefore:
# moles of Cl = 0.004478 moles
At wt of Cl = 35.45 g/mol
Mass%(Cr) = 100 - 14.50=85.5%