Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
----------------------------------
Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
Answer:
see explanation...
Explanation:
Mg⁺²-24 Co⁺³-60 Clˉ-35
Protons (p⁺) 12 27 17
Neutrons (n⁰) 12 33 18
Electrons (eˉ) 10 24 18
(c) (b) (a)
12/2 : 12/2 : 10/2 27/3 : 33/3 : 24/3 #n⁰ = 18
6 : 6 : 5 9 : 11 : 8 #eˉ = 18
The number of C2H5OH in a 3 m solution that contain 4.00kg H2O is calculate as below
M = moles of the solute/Kg of water
that is 3M = moles of solute/ 4 Kg
multiply both side by 4
moles of the solute is therefore = 12 moles
by use of Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 12 moles
=12 moles/1 moles x 6.02 x10^23 = 7.224 x10^24 molecules
Answer:
A. Compounds
Explanation:
Firstly, don't let the word "pure" confuse you; this is pretty much irrelevant to the question.
The key to answering this question lies in "substance" and "more than one type of element chemically bonded."
Something you ought to memorize is that a substance is either an element or compound. Therefore, you can eliminate choices B./C.
Next, use the part of the definition that says "more than one type of element chemically bonded" to conclude that it's a compound. Not only is this the definition of a compound, but the fact that is says <em>more than one type of element</em> should automatically tell you that it is different from just a regular element (Choice D.).
