The concentration of A will be <em>0.34 mol·L⁻¹</em> after 60 min.
In a first-order reaction, the formula for the amount remaining after <em>n</em> half-lives is
![\text{[A]} = \frac{\text{[A]}_{0}}{2^{n}}\\](https://tex.z-dn.net/?f=%5Ctext%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%7B2%5E%7Bn%7D%7D%5C%5C)
If 
∴ ![\text{[A]} = \frac{\text{2.7 mol/L}}{2^{3.0}} = \frac{\text{2.7 mol/L}}{8.0} = \textbf{0.34 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B%5Ctext%7B2.7%20mol%2FL%7D%7D%7B2%5E%7B3.0%7D%7D%20%3D%20%5Cfrac%7B%5Ctext%7B2.7%20mol%2FL%7D%7D%7B8.0%7D%20%3D%20%5Ctextbf%7B0.34%20mol%2FL%7D)
Answer:
H₂SO₄
Explanation:
We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.
Step 1: Calculate the total mass of the compound
Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g
Total mass = 23.139 g
Step 2: Determine the percent composition.
H: (0.475g/23.139g) × 100% = 2.05%
S: (7.557g/23.139g) × 100% = 32.66%
O: (15.107g/23.139g) × 100% = 65.29%
Step 3: Divide each percentage by the atomic mass of the element
H: 2.05/1.01 = 2.03
S: 32.66/32.07 = 1.018
O: 65.29/16.00 = 4.081
Step 4: Divide all the numbers by the smallest one
H: 2.03/1.018 ≈ 2
S: 1.018/1.018 = 1
O: 4.081/1.018 ≈ 4
The empirical formula of the compound is H₂SO₄.
Answer:
It's B !
Explanation:
Formulas. The molecular formula for glucose is C6H12O6. This means that there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms bonded together to make one molecule of glucose.
Hope this helps!!
An example of an element is C) silver
