Group 18 is known as the Noble/ Inert Gases
Answer:
Explanation:
<u>1) Find the z-scores:</u>
a) z-score for 22.6 inches length
- z = [ 22.6 - 20 ] / 2.6 = 1.00
b) z-score for 17.4 inches length
- z = [ 17.4 - 20 ] / 2.6 = - 1.00
<u>2) Probability</u>
Then, you have to find the probability that the length of an infant is between - 1.00 and 1.00 standards deviations (σ) from the mean (μ).
That is a well known value of 68%, which is part of the 68-95-99.7 empirical rule.
The most exact result is obtained from tables and is 68.26%:
- 1 - P (z ≥ 1.00) - P (z ≤ - 1.00) = 1 - 0.1587 - 0.1587 = 0.6826 = 68.26%
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
