Explanation:
Sodium Chloride or NaCl is made up of two elements, sodium (or Na) and chlorine (or Cl). A molecule of sodium chloride, NaCl, consists of one atom each of sodium and chlorine. Hence, each molecule of NaCl has 2 atoms total.
Answer:
68133080.02 g
Explanation:
I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.
Now, if 1 mole of a gas occupies 22.4 L
x moles of air occupies 52,681,428.8 Liters
x = 1 * 52,681,428.8 /22.4
x = 2351849.5 moles of air
Now, number of moles = mass/ molar mass
but molar mass = 28.97 g/mol
2351849.5 = mass/28.97
mass = 2351849.5 * 28.97
mass = 68133080.02 g
Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:

Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>
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Spore formation is a form of asexual reproduction used by mushrooms and molds.
During budding, the offspring grows from the body of the parent.
Fragmentation is a form of asexual reproduction that must be followed by regeneration.
Explanation:
Asexual reproduction is the type of reproduction where the gamete formation and fusion have no relevance or existence. It functions on the process of somatic cell division via mitosis and the offsprings are identical to their parents.
The spore formation occurs in fungi through sporangia, bursting open to shed spores, forming into a new young ones. Budding occurs out as an outgrowth of the parent and attains maturity and separates. Fragmentation is the process where the parents fall apart into pieces and regeneration follows.
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.