Answer:
Explanation:
The Solar System is the gravitationally bound system of the Sun and the objects that orbit it, either directly or indirectly. Of the objects that orbit the Sun directly, the largest are the eight planets, with the remainder being smaller objects, the dwarf planets and small Solar System bodies
Answer:
B. Group 17
Explanation:
The element is Bromine. Its electron configuration is:
1s2 2s2p6 3s2p6d10 4s2p5
You can see that its last orbital is 4p^5 for 5 electrons in the 4p orbital.
Answer:
For every temperature you use the formula given by the question.
at 3°C you have:
Reaction rate = (1000/0.2)/138.5 = 36 mg/L/sec
at 24°C you have:
Reaction rate = (1000/0.2)/34.2 = 146 mg/L/sec
at 40°C you have:
Reaction rate = (1000/0.2)/26.3 = 190 mg/L/sec
at 65°C you have:
Reaction rate = (1000/0.2)/14.2 = 352 mg/L/sec
Explanation:
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
