Answer:
d Rubidium
Explanation:
The atomic radius of an atom is the distance from the center of the nucleus to its outermost electron.
The atomic radius of elements varies in the periodic table, it increases as you go down in a group and decreases along the period from left to right.
All the elements listed: Hydrogen, Sodium, Lithium and Rubidium belong to the same group in the periodic table (group 1), Since atomic radius increases from top to bottom in a group, Rubidium has the largest atomic radius.
The answer that complete the sentence is "Arrhenius base".
The complete sentence is: A substance that increases the concentration of hydroxide ions in aqueous solution is an Arrhenius base.
Arrhenius' acid and base model says that an acid is a substance that contains hydrogen and ionizes in aqueous solution producing protons (H+) and a base is a substance that cointains hydroxide and release it in aqueous solution.
Answer:
b. HCOOH/ NaHCOO.
Explanation:
A buffer system may be formed in one of two forms:
- A weak acid with its conjugate base.
- A weak base with its conjugate acid.
Chose the pairs below that you could use to make a buffered solution.
a. HCI/NaOH. NO. HCl is a strong acid and NaOH is a strong base.
b. HCOOH/ NaHCOO. YES. HCOOH is a weak acid and HCOO⁻ (coming from NaHCOO) is its conjugate base.
c. HNO₂/H₂SO₃. NO. Both are acids and they are unrelated to each other.
d. NaNO₃/ HNO₃. NO. HNO₃ is a strong acid.
Answer:
The mass of the neon gas m = 1.214 kg
Explanation:
Pressure = 3 atm = 304 k pa
Volume = 0.57 L = 0.00057 
Temperature = 75 °c = 348 K
Universal gas constant = 0.0821 
We have to change the unit of this constant. it may be written as
Universal gas constant = 8.314 
Gas constant for neon =
= 0.41 
From ideal gas equation,
P V = m R T ------- (1)
We have all the variables except m. so we have to solve this equation for mass (m).
⇒ 304 ×
× 0.00057 = m × 0.41 × 348
⇒ 173.28 = 142.68 × m
⇒ m = 1.214 kg
This is the mass of the neon gas.
Answer:
6.48 L
Explanation:
From the question,
Applying
PV/T = P'V'/T'......................... Equation 1
P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.
make V' the subject of the equation
V' = PVT'/P'T......................... Equation 2
Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm
Substitute these values into equation 2
V' = (4.5×1×253)/(0.6×293)
V' = 1138.5/175.8
V' = 6.48 L