This question is incomplete, the complete question is;
A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.
Answer:
the final pH is 3.187
Explanation:
Given the data in the question;
Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol
Initial moles of NO2- = mass/molar mass = 5.86/68.995 = 0.0849336 mol
Moles of KOH added = 10.97/1000 × 1.63 = 0.0178811 mol
so
HN02 + KOH → NO2- + H2O
moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol
Moles of NO2- = 0.0849336 + 0.0178811 = 0.1028147 mol
Now,
pH = pka + log( [NO2-]/[HNO2])
pH = -log ka + log( moles of NO2- / moles of HNO2 )
we substitute
pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147 / 0.1418589 )
pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )
pH = 3.337242 + (-0.1398 )
pH = 3.187
Therefore, the final pH is 3.187
Answer:
The third option
Explanation:
If people have found more volcanic rock layers in the past that would mean that volcanic activity was more common in the past.
Answer:
Litmus paper is one type of acid-base indicator. The paper is imbued with dye derived from lichens that change color in response to the presence of an acid or base. ... Red paper is used to detect alkaline pH and will turn a shade of blue in the presence of a basic solution.
Answer:
Explanation:
To interpret this pedigree, let’s start with information that we already know:
Brown is recessive, which means brown individuals must have the phenotype BB. In this pedigree, brown individuals are filled in.
Black is dominant, which means black individuals must have at least one B allele. Their phenotype could be either BB or BB. In this pedigree, black individuals are not filled in.
Figure 5 shows the same pedigree, but with information about the individual’s phenotype filled in.
The shaded individual, who is a brown female puppy, must have the phenotype BB. If she had any B alleles, she would be black because the black allele is dominant over the brown allele.
In order for the brown puppy to have the phenotype BB, she must have gotten two “b” alleles: one from each of her parents. We know that her parents are both black (because they are unshaped), which means they must have a least one “B” allele. This means that both parents must be heterogeneous: BB.
The three black puppies must have at least one “B” allele in order for them to be black in color. However, we can’t tell whether they are homologous dominant (BB) or heterogeneous (BB) since both of those phenotype would result in black color. One way to represent this on a pedigree is B-, meaning that the second allele could be either B or b.
Answer:
111 L
Explanation:
Calculation of moles of hydrogen gas:-
Mass of 
= 18.6 g
Molar mass of = 2.01588 g/mol
According to the given reaction:-
2 moles of hydrogen gas on reaction produces one mole of acetic acid gas.
So,
1 mole of hydrogen gas on reaction produces 
mole of acetic acid gas.
Also,
9.23 mole of hydrogen gas on reaction produces 
mole of acetic acid gas.
Moles of acetic acid gas = 4.615 moles
Given that:
Temperature = 35 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (35 + 273.15) K = 308.15 K
n = 4.615 moles
P = 1.05 atm
V = ?
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
1.05 atm × V = 4.615 moles ×0.0821 L atm/ K mol × 308.15 K
<u>⇒V = 111 L</u>
