Answer:
the molecular formula for the gas is NO₂
Explanation:
since it contains
Nitrogen = n → 30.45%
Oxygen = o → 69.55%
and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen
Also we know that the proportion of oxygen over nitrogen is
proportion of oxygen over nitrogen = moles of oxygen / moles of nitrogen
since
moles = mass / molecular weight
then for a sample of 100 gr of the unknown gas
mass of oxygen = 69.55%*100 gr = 69.55 gr
mass of Nitrogen = 30.45%*100 gr = 30.45 gr
proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) = (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N
therefore there are 2 atoms of oxygen per atom of nitrogen
thus the molecular formula for the gas is:
NO₂
Answer:
20 pig callers
Explanation:
Given that:
A pig caller produced intensity level of a sound = 107 dB
To find how many pig callers required to generate an intensity level of 120 dB;
we have:
120 dB - 107 dB = 13 dB
Taking the logarithm function;

where;
= initial intensity


I = 19.95
I ≅ 20 pig callers
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct