Given :
A 10.99 g sample of NaBr contains 22.34% Na by mass.
To Find :
How many grams of sodium does a 9.77g sample of sodium bromine contain.
Solution :
By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.
Therefore , percentage of Na by mass in NaBr will be same for every amount .
Percentage of Na in 9.77 g NaBr is 22.34 % too .
Gram of Na =
.
Hence , this is the required solution .
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in
of ethane gas
And, as we know that
1 mole of ethane molecule contains
molecules of ethane
2.869 moles of ethane molecule contains
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is,
molecule.
Answer:
c) atomic number / alkaline earth metals/ and halogens
Explanation:
Elements of modern periodic table are arrang in atomic number; for its electron configuration and its chemical properties. This arrangement shows periodic trends.
Alkaline earth metals are a group of elements that are located in group 2 of the Periodic Table and are the following: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba) and Radio (Ra).
The Halogens are the chemical elements that form group 17 (XVII A, previously used) or group VII A of the periodic table: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At) and teneso (Ts)
Answer: 568g/mol
Explanation:
It should be noted that there are 40 atoms of carbon in lycopene.
Since mass of 1 carbon = 12g/mol
Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol
Let the molar mass of lycopene be represented by x.
Therefore the molar mass of carbon = x × mass percent of carbon in lycopene
x × 84.49% = 480g/mol
x × 0.8449 = 480g/mol
x = 480/0.8449
x = 568g/mol
The molar mass of lycopene is 568g/mol
Answer:
1. 0.073L
2. 0.028L
3. 0.014L
Explanation:
The volume for the different solutions are obtained as shown below:
1. Mole = 0.53mol
Molarity = 7.25M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.53/7.25
Volume = 0.073L
2. 0.035mol from a 1.25M
Mole = 0. 035mol
Molarity = 1.25M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.035/1.25
Volume = 0.028L
3. Mole = 0.0013mol
Molarity = 0.090M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.0013/0.090
Volume = 0.014L