Answer:
C
Explanation:
I got this question right on my test.
Determine the number of electrons in an Cl ion with a charge of -3
Answer:
There are 20 electrons.
Explanation:
Chlorine has 17 electrons in its atom to begin with (you can tell from the proton number in the periodic table).
A charge of -3 means that the chlorine atom gained 3 electrons, and now needs to lose 3 electrons to return to its atom configuration.
Are sperm and egg cells exact copies of the plant cell
Answer:
No
Explanation:
thats scientifically impossible
Potassium nitrate, KNO3 , has a molar mass of 101.1 g/mol. In a constant-pressure calorimeter, 13.3 g of KNO3 is dissolved in 213 g of water at 23.00 °C . KNO3(s)−→−−H2OK+(aq)+NO−3(aq) The temperature of the resulting solution decreases to 19.60 °X . Assume that the resulting solution has the same specific heat as water, 4.184 J/(g·°C) , and that there is negligible heat loss to the surroundings. How much heat was released by the solution
Answer:
[tex]Q_{sln}=-3219.25 J[/tex]
Explanation:
Hello there!
In this case, for this calorimetry problem, it is possible for us to infer that the heat of the reaction of dissolution of KNO3 is absorbed by the solution composed by the former and water so that we can write:
[tex]Q_{rxn}=-Q_{sln}[/tex]
Thus, given the mass, specific heat and temperature of the solution, we plug in the data to obtain the heat absorbed, by the reaction:
[tex]Q_{rxn}=-m_{sln}C_{sln}\Delta T _{sln}\\\\Q_{rxn}=-(13.3+213)g(4.184\frac{J}{yg\°C} )(19.60-23.00)\°C\\\\Q_{rxn}=3219.25J[/tex]
Also, we can say the the heat released by the solution was -3219.25 J.
Best regards!
You are performing an acid-base neutralization reaction in the laboratory to determine the concentration of an unknown base. You are supposed to titrate it with a monoprotic acid, but your lab partner accidentally fills your buret with sulfuric acid, a diprotic acid, with the same concentration as the acid called for in the experiment. How will the volume of diprotic acid compare to the volume of monoprotic acid you would have used
Answer:
Volume is reduced to half
Explanation:
Acid base titration are commonly used reactions in a lab, and are ofter used to get pH or different kind of solutions.
The neutralization of an acid base reaction is reached, when the solution (having added an indicator previously) changes its original color. chemically speaking, this occurs when the number of moles of the acid and the base are balanced and equal. In other words the following:
n₁ = n₂ (1)
This expression can also be expressed in function of concentration and volume:
M₁V₁ = M₂V₂ (2)
From here, solving for V₁:
V₁ = M₂V₂ / M₁
Now, this expression is true only when we have the same kind of substance that can lose or gain the same number of hydrogens.
Lets suppose that we have as base NaOH (Monoprotic base) and HCl (monoprotic acid), the titration reaction would be:
NaOH + HCl --------> NaCl + H₂O
As both of the species are monoprotic, the number of moles are the same when they reach the equilibrium, so, expression (2) can be used, and calculate volume or concentrations.
However, in this case, a partner made a mistake and use a diprotic acid, in this case, H₂SO₄, In this case, things chance because H₂SO₄ is diprotic, meaning that we need to dissociate two hydrogens in equilibrium, therefore, expression (2) would be something like this.
Acid: 1; Base: 2
H₂SO₄ + 2NaOH ------> Na₂SO₄ + H₂O
nH₂SO₄ = n₁ = 1
nNaOH = n₂ = 2
n₁/n₂ = 1/2
2n₁ = n₂ (3)
Writting this, in function of concentration and volume, it would be:
2M₁V₁ = M₂V₂ (4)
From here, if we solve for the volume of the acid (V₁):
V₁ = M₂V₂ / 2M₁
Therefore, according to this expression, we can see that the volume required of the acid would be half the volume required of the monoprotic acid. For example, if we need 50 mL of Chloridic acid to reach the equivalence point with NaOH, then, with H₂SO₄ it will only need 25 mL. This, of course, assuming that concentrations are the same, and volume of the base used, the same.
Hope this helps
Consider the following data on some weak acids and weak bases
acid
Ka
name formula
acetic acid
HCH3CO2
1.8 x10−5
hydrocyanic acid
HCN
4.9 x 10−10
base
Kb
name formula
pyridine
C5H5N
1.7 x 10−9
ammonia
NH3
1.8 x 10−5
Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on.
a. 0.1M NaCH3CO2
b. 0.1M NH4Br
c. 0.1M NaBr
d. 0.1M KCN
Answer:
b < c < a < d
Explanation:
The weak acid with the lowest pKa will be the most acidic. In the other way, the conjugate base which the acid is weak will be strong.
The weak base with the lowest pKb will be the most basic. And the conjugate base of the weak base will be a strong acid.
Ka Acetic acid = 1.8x10-5
Ka HCN = 1.9x10-10
Kb pyridine = 1.7x10-9
Kb NH3 = 1.8x10-5
NH4Br is the conjugate base of a weak base. That means is a strong acid.
NH4Br has the lowest pH
NaBr is the conjugate base of a strong acid, HBr. That means NaBr is neutral
The most basic between the conjugate base of the acetic acid, NaCH3CO2 and KCN is KCN because the acetic acid is the stronger acid regard to HCN.
The rank is:
NH4Br < NaBr < NaCH3CO2 < KCN
b < c < a < d