Answer:
Hydrogen ion concentration is 2.773E-4Mole per DM3
Hydroxide ion concentration is 4.093E-11
It is an acid
Explanation:
PH(H3O)=-LogH3O
PH=3.557
H3O=antilog of -3.557
antilog of -3.557=antilog (0.443 -4)
antilog0.443 ×antilog -4=2.773E-4mole per DM3.
(H3O)(OH)=10^-14
H30=2.773×10^-14
OH= 1.00E-14/2.773E-4
=4.093E-11
it is an acid since the concentration of hydrogen ion is more than the hydroxide ion
Answer:
-Being in the service of quality, safety, designing and problem solving.
-It plays an importnat part in our lives too, to measure any surface, object, etc.
Explanation:
Measurement is perhaps one of the most fundamental concepts in science. Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories.
Answer:
Limiting: Lab covers. Books produced: 75. Excess amounts: In explanation
Explanation:
Amount that can be produced with just lab covers: 75 books (150/2)
Amount that can be produced with just lined paper: 150 books (7500/50)
Amount that can be produced with graph paper: 120 (3000/25)
Amount that can be produced with staples: Approximately 83 (250/3)
As we can see, the lab covers are limiting as they can only produce 75 books. So, we can only make 75 books.
You will have 3,750 lined paper left over (or 75 books)
(150-75=75 , 75*50=3,750)
You will have 1,125 graph paper left over (or 45 books)
(120-75=45 , 75*25=1,875 , 3000-1875=1125)
And then you will have approximately 25 staples left over (or 8 books)
(83-75=8 , 8*3=225, 250-225=25)
(Hopefully this is correct, I apologize if I messed up)
Answer:
Option A) Exothermic Reaction
Explanation:
In exothermic reaction, the energy is released. The reactants are at high energy while the products are at low energy as shown in the graph.
Answer:
The molar composition of the equilibrium mixture is
NO: 0.338 = 33.8%
Br2: 0.169 = 16.9%
NOBr: 0.493 = 49.3%
Explanation:
The reaction of nitrosyl bromide formation can be written as
To form 1 mol of NOBr, we need 1 mol of NO and 0.5 mol of Br2.
A sample of 0.0524 mol NO with 0.0262 mol Br2 gives an equilibrium mixture containing 0.0311 mol NOBr.
Then, to form 0.0311 mol NOBr, were needed 0.0311 mol NO and 0.01555 mol of Br2.
The amount of NO that stay in the same form is (0.0524-0.0311)=0.0213 mol NO.
The amount of Br2 that stay in the same form is (0.0262-0.01555)=0.01065 mol Br2.
The total amount of mol is (0.0213 mol NO + 0.01065 mol Br2 + 0.0311 mol NOBr) = 0.06305.
The molar composition is
NO: 0.0213/0.06305 = 0.338 = 33.8%
Br2: 0.01065/0.06305 = 0.169 = 16.9%
NOBr: 0.0311/0.06305 = 0.493 = 49.3%
