Answer:
0.5188 M or 0.5188 mol/L
Explanation:
Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.
***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".
The formula for molarity is:
n = moles (unit mol)
V = volume (unit L)
<u>Find the molar mass (M) of potassium hydroxide.</u>
<u>Calculate the moles of potassium hydroxide.</u>
Carry one insignificant figure (shown in brackets).
<u>Convert the volume of water to litres.</u>
Here, carrying an insignificant figure doesn't change the value.
<u>Calculate the concentration.</u>
<= Keep an insignificant figure for rounding
<= Rounded up
<= You use the unit "M" instead of "mol/L"
The concentration of this standard solution is 0.5188 M.
Answer:
a) Addition reaction, is your answer
Answer:
C. Lose three electrons to have a full outer shell
Explanation:
Al is in Group 13 of the Periodic Table, so it has three valence electrons.
It must either lose three electrons or gain five to achieve a stable octet.
It is easier to lose three electrons than it is to gain five, so Al loses three electrons.
D. is wrong, for the same reason.
A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.
C. is wrong. Al is a metal, so it will lose electrons in a reaction.
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.
Answer:
a) v = 1497.2 cm^-1
b) v = 1465 cm^-1
Explanation:
In the attached image is the procedure explained to reach the answer.
