Answer:
i think the answer is B cus i think of that
Sulfur trioxide (SO3) is a chemical compound that is a significant pollutant in gaseous form as it is involved in the production of acid rain.
Industrially, sulfur trioxide is an important precursor to sulfuric acid and is formed from the reaction between sulfur dioxide (SO2) and oxygen gas (O2) as shown in the chemical equation below.
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
Answer:
2,760 grams NaCl
Explanation:
To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.
2 Na + Cl₂ --> 2 NaCl
Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl) = 58.44 g/mol
47.2 moles Na 2 moles NaCl 58.44 grams
---------------------- x --------------------------- x ------------------------- =
2 moles Na 1 mole NaCl
= 2,758.368 grams NaCl
= 2,760 grams NaCl
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.

Temperature = 
As molar mass of
is 32 g/mol. Hence, the number of moles of
are calculated as follows.

Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.