Answer is: d). 1-propanol. Because between two molecules of 1-propanol can for form hydrogen bond, between hydrogen and oxygen in hydroxy group.
1) Chemical formula for propanal is CH₃-CH₂-CH=O. Propanal is a saturated three carbon aldehyde (have a carbonyl center).
2) Chemical formula for propane is CH₃-CH₂-CH₃. Propane is a three carbon alkane (acyclic saturated hydrocarbon).
3) Chemical formula for propanone is (CH₃)₂-C=O. Propanone or acetone is he simplest and smallest ketone.
4) Chemical formula for propanol is CH₃-CH₂-CH₂-OH. 1-propanol is a primary alcohol.
Hydrogen bond is
an electrostatic attraction between two polar groups that occurs when
a hydrogen atom (H), covalently bound to a
highly electronegative atom such as flourine (F), oxygen (O) and
nitrogen (N) atoms.
Answer:
0.007756M KHP; 0.0009695 moles NaOH reacts; 0.03488M NaOH.
Explanation:
Potassium hydrogen phthalate, KHP, is a salt used as standard to determine concentration of basic solutions as NaOH solutions.
To find molarity of the KHP solution we need to convert mass of KHP to moles and divide this in 125mL = 0.125L:
<em>Moles KHP -Molar mass: 204.22g/mol-:</em>
0.198g * (1mol / 204.22g) = 0.0009695 moles KHP
<em>Molarity:</em>
0.0009695 moles KHP / 0.125L = 0.007756M KHP
NaOH reacts with KHP as follows:
NaOH + KHP → Na⁺ + KP⁻ + H₂O
That means 1 mole of NaOH reacts per mole of KHP, that means moles of NaOH that reacts are = Moles of KHP added:
0.0009695 moles NaOH reacts
In 27.80mL = 0.02780L the NaOH was contained. Molarity is:
0.0009695 moles NaOH / 0.02780L =
0.03488M NaOH
D is the answer here look
Atomic weight of an element can be calculated as follows:
average atomic weight =
(atomic weight of first isotope)(its percentage of abundance) +
(atomic weight of second isotope)(its percentage of abundance)
average atomic weight = 98.225 amu
atomic weight of first isotope = 97.780 amu
its percentage of abundance = 1 - 0.417 = 0.583
atomic weight of second isotope = ??
its percentage of abundance = 0.417
So, just substitute in the above equation to get the atomic weight of the second isotope as follows:
98.225 = (97.78)(0.583) + (mass2)(0.417)
atomic weight of second isotope = 98.847 amu
Answer:
I) 0.0585 M
ii)6.2 g dm-3
Explanation:
The reaction equation is given as;
Na2CO3(aq) +2HCl(aq)------> 2NaCl(aq) + CO2(g) +H2O(l)
Concentration of acid CA= 0.3 M
Volume of acid VA= 3.9 cm^3
Concentration of base CB= the unknown
Volume of base VB= 10 cm^3
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB=CBVBNA
CB= CAVANB/VBNA
substituting values;
CB= 0.3 × 3.9 × 1/ 10.0 × 2
CB= 0.0585 M
ii) mass concentration= molar concentration × molar mass
Molar mass of Na2CO3= 106 gmol-1
Mass concentration= 0.0585 × 106 = 6.2 g dm-3