Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J


ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Answer:
They both have the same number of atoms
Explanation:
The number that indicates the amount of particles in a compound is the Avogadro's number (NA).
It does not matter the mass of compound we have, If we have 1 mol we will be sure that we are talking about 6.02×10²³ particles
6.02×10²³ represents the amount of atoms in twelve grams of 12-pure carbon and it is considered a reference to measure the amount of all kinds of substances present in a given system.
Answer:
The mass percent of potassium is 39%
Option C is correct
Explanation:
Step 1: Data given
Atomic mass of K = 39.10 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate molar mass of KHCO3
Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0
Molar mass KHCO3 = 100.12 g/mol
Step 3: Calculate mass percent of potassium (K)
%K = (atomic mass of K / molar mass of KHCO3) * 100%
%K = (39.10 / 100.12) * 100%
%K = 39.05 %
The mass percent of potassium is 39%
Option C is correct
Answer: gasoline, water, sea water, chloroform and mercury so B
Explanation: