A.
to defend against infection
B.
to provide structural support
C.
to carry genetic information
D.
to process information
(2) 0.1 M KOH (4) 0.1 M H2SO4
Answer: (2) 0.1 M KOH
Explanation:
Thymol blue is an universal indicator. Thymol blue has a blue color when in basic conditions (pH > 7), a green color in neutral conditions (pH =7), and a yellow color in acidic conditions (pH < 7).
pH or pOH is the measure of acidity or alkalinity of a solution. pH is calculated by taking negative logarithm of hydrogen ion concentration.
0.1 M , 0.1 M and 0.1 M are all acidic in nature.
Thus thymol blue will appear blue in basic 0.1 M KOH.
Answer:
24.03 J/mol.ºC
Explanation:
For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.
-Qal = Qw + Qc (minus signal represents that the heat is lost)
-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc
Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC
-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)
1971.15Cal = 1699.10 + 58.22
1971.15Cal = 1757.32
Cal = 0.89 J/g.ºC
The molar mass of Al is 27 g/mol
Cal = 0.89 J/g.ºC * 27 g/mol
Cal = 24.03 J/mol.ºC
Answer
1.60217662 × 10-19 coulombs
Explanation:
.
FALSE
Answer:
The answer is True
Explanation:
Cyclic hydrocarbons are organic compounds that consist of carbon and hydrogen atoms that bond to form rings or cyclic structures. The simplest are cycloalkanes, hydrocarbons formed by a closed carbon chain with all the simple bonds.
There is a very special group of cyclic hydrocarbons called aromatic hydrocarbons. Its name is due to the fact that the first to be discovered had pleasant smells. They all start from a basic structure that can be repeated many times. An example of this type of hydrocarbon is bancene.
The balanced combustion reaction for butane is;
2C4H10 + 13O2 → 8CO2 + 10H2O
Molar mass of water = 18.02 g/mol
Molar mass of oxygen = 32 g/mol
2.46g H2O(1 mol H2O/18.02g)(13 mol O2/10 mol H2O)(32g O2/1 mol O2) = 5.68g O2