The answer is temperature
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:
2Mg + O₂ ⟶ 2MgO
Explanation:
Step 1. Start with the most complicated-looking formula (O₂?).
Put a 1 in front of it.
Mg + 1O₂ ⟶ MgO
Step 2. Balance O.
We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.
Mg + 1O₂ ⟶ 2MgO
Step 3. Balance Mg.
We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.
2Mg + 1O₂ ⟶ 2MgO
Every formula now has a coefficient. The equation should be balanced. Let’s check.
<u>Atom</u> <u>On the left</u> <u>On the righ</u>t
Mg 2 2
O 2 2
All atoms are balanced.
The balanced equation is
2Mg + O₂ ⟶ 2MgO
Answer:
there are 3 significant figures
Explanation:
do not count the first 2 zeros. only the nimbers after the zero
Answer:
10.8 ml
Explanation:
The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.
See attached file