We can use a variety of formulas to determine our answers here.
Our formula for pOH is -log(mol), and we can plug it in as -log(0.010). Take note that OH- is a base, not an acid.
So, the pOH of OH- is 2.
To find pH we can set up this simple equation:
pH + pOH = 14
All we need to do is subtract 2 from 14, therefore the pH is 12.
This makes sense since acids range in the pH of 1-6, and we are dealing with a base. Hope I could help!
Answer:
C. Y & Z
Explanation:
V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).
Thus, Y and Z make covalent.
Answer:
330 mL of (NH₄)₂SO₄ are needed
Explanation:
First of all, we determine the reaction:
(NH₄)₂SO₄ + 2NaOH → 2NH₃ + 2H₂O + Na₂SO₄
We determine the moles of base:
(First, we convert the volume from mL to L) → 62.6 mL . 1L/1000 mL = 0.0626L
Molarity . volume (L) = 2.31 mol/L . 0.0626 L = 0.144 moles
Ratio is 2:1. Therefore we make a rule of three:
2 moles of hydroxide react with 1 mol of sulfate
Then, 0.144 moles of NaOH must react with (0.144 .1) /2 = 0.072 moles
If we want to determine the volume → Moles / Molarity
0.072 mol / 0.218 mol/L = 0.330 L
We convert from L to mL → 0.330L . 1000 mL/1L = 330 mL
Answer:- 0.800 moles of the gas were collected.
Solution:- Volume, temperature and pressure is given for the gas and asks to calculate the moles of the gas.
It is an ideal gas law based problem. Ideal gas law equation is used to solve this. The equation is:
PV=nRT
Since it asks to calculate the moles that is n, so let's rearrange this for n:
V = 19.4 L
T = 17 + 273 = 290 K
P = 746 mmHg
we need to convert the pressure from mmHg to atm and for this we divide by 760 since, 1 atm = 760 mmHg
P = 0.982 atm
R = 
Let's plug in the values in the equation to get the moles.
n = 0.800 moles
So, 0.800 moles of the gas were collected.
Answer:
The order of reactivity towards electrophilic susbtitution is shown below:
a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene
b. p-cresol>p-xylene>toluene>benzene
c.Phenol>propylbenzene>benzene>benzoic acid
d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene
Explanation:
Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.
For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.
The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.
Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.
