Answer: False
Explanation:
4C2H6 + 7O2 --> 6H2O + 4CO2
8 Carbons on the reactant side, but 4 Carbons on the Product
24 Hydrogens on Reactant, 12 H on product
But Oxygen is balanced, 14 on each side
Answer:
Option B. 3.0 M
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity can simply be defined as the mole of solute per unit litre of the solution. Mathematically, it can be expressed as:
Molarity = mole of solute /Volume of solution
With the above formula, we can obtain the molarity of the solution as follow:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity = mole of solute /Volume of solution
Molarity = 9 / 3
Molarity = 3 mol/L = 3.0 M
Thus, the molarity of the solution is 3 M
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
You can answer this question by only searching the element in the periodic table.
The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).
The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
