<u>Answer:</u> The mass of nitrogen that is present for given amount of aluminium is 566.22 g.
<u>Explanation:
</u>
We are given:
Mass of aluminium = 364 grams
The chemical formula of aluminium nitrate is 
Molar mass of nitrogen = 14 g/mol
Molar mass of aluminium = 27 g/mol
In 1 mole of aluminium nitrate, 27 grams of aluminium is combining with 42 grams of nitrogen.
So, 364 grams of aluminium will combine with = 
of nitrogen.
Hence, the mass of nitrogen that is present for given amount of aluminium is 566.22 g.
Answer:
B) a helium nucleus moving at a velocity of 1000 mph
Explanation:
According to the De Broglie relation
λ= h/mv
h= planks constant
m= mass of the body
v= velocity of the body.
As we can see from De Broglie's relation, the wavelength of matter waves depends on its mass and velocity. Hence, a very small mass moving at a very high velocity will have the greatest De Broglie wavelength.
Of all the options given, helium is the smallest matter. A velocity of 1000mph is quite high hence it will have the greatest De Broglie wavelength.
In given data:
maximum absorption wavelength λ = 580 nm = 580 x 10⁻⁹ m
write the equation to find the crystal field splitting energy:
E = hC / λ
Here, E is the crystal field splitting energy, h = 6.63 x 10⁻³⁴ J.sec is Planck's constant and C = 3 x 10⁸ m/sec is speed of light.
substitute in the equation above:
E = (6.64 x 10⁻³⁴ x 3 x 10⁸) / (580 x 10⁻⁹) = 3.43 x 10⁻¹⁹J
This crystal field splitting energy is for 1 ion.
Number of atoms in one mole, NA = 6.023 x 10²³
to calculate the crystal field splitting energy for one mole:
E(total) = E x NA
= (3.43 x 10⁻¹⁹) x (6.023 x 10²³) = 206 kJ/ mole
Answer:
29.98kg
Explanation:
12.0 gallons * (3.78541178 liters/gallon) * (1000 mL/liter) * (0.66 g/mL) * (1 kg/1000 g) = 29.98 kg
718.65 degrees is the initial temperature of the zinc metal sample.
Explanation:
Data given:
mass of zinc sample = 2.50 grams
mass of water = 65 grams
initial temperature of water = 20 degrees
final temperature of water = 22.5 degrees
ΔT = change in temperature of water is 2.50 degrees
specific heat capacity of zinc cp= 0.390 J/g°C
initial temperature of zinc sample = ?
cp of water = 4.186 J/g°C
heat absorbed = heat released (no heat loss)
formula used is
q = mcΔT
q water = 65 x 4.286 x 2.5
q water = 696.15 J
q zinc = 2.50 x 0.390 x (22.50- Ti)
equating the two equations
696.15 = - 22.50+ Ti
Ti = 718.65 degrees is the initial temperature of zinc.
