Answer:
Stoichiometric Coefficients
The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element. Understanding this is essential to solving stoichiometric problems
Explanation:
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
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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 %
Explanation:
Balanced chemical reaction equation will be as follows.
In human body, the neutral iron changes into 
(aq) cation. There will be an oxidation-half reaction and a reduction-half reaction. Equations for this reaction are as follows.
Oxidation: 2Fe^{2+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + 2e^{-}[/tex] .... (1)
Reduction: 
...... (2)
On adding both equation (1) and (2), the overall reaction equation will be as follows.
Therefore, neutral iron is a part of Heme - b group of Hemoglobin and in an aqueous solution it dissolutes as a part of Heme group. Hence, then it becomes an cation.
Answer:
Solution is 0.28 M
You can also say, [NaCl] = 0.28 mol/L
Explanation:
As you have a solute mass and the solution's volume, you may find the molarity concentration of solution.
Molarity specifies the moles of solute in 1 L of solution
We convert the volume of solution to L → 350 mL . 1L / 1000 mL = 0.350L
We convert the mass of solute to moles → 5.80 g . 1mol / 58.45 g = 0.0992 moles
Molarity (mol/L) = 0.0992 mol /0.350L = 0.28M
