a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
Answer:
C.
Explanation:
Iron is most likely to form a precipitation reaction.
When energy transforms into mass, the amount of energy does not remain the same. When mass transforms into energy, the amount of energy also does not remain the same. However, the amount of matter and energy remains the same. ... You would weigh much less on the Moon because it is only about one-sixth the mass of Earth. So the answer is D
Answer:
See explanation below
Explanation:
To get this, we need to apply the general expression for half life decay:
N = N₀e(-λt) (1)
Where:
N and N₀ would be the final and innitial quantities, in this case, masses.
t: time required to decay
λ: factor related to half life
From the above expression we need λ and t. To get λ we use the following expression:
λ = t₁₂/ln2 (2)
And we have the value of half life, so, replacing we have:
λ = 8.04 / ln2 = 11.6
Now, we can replace in (1) and then, solve for t:
0.75 = 40 exp(-11.6t)
0.75 / 40 = exp(-11.6t)
ln(0.01875) = -11.6t
-3.9766 = -11.6t
t = -3.9766 / -11.6
<h2>
t = 0.34 days</h2><h2>
</h2>
