Answer:
16.0%.
Explanation:
Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.
V % of ethanol = (volume of ethanol / volume of the solution) x 100.
volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.
∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.
Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas
Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988
Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol
now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1
1 C2H6 + 1 O2 -> products
this means that for 1 mol of C2H6, 1 mol of O2 has to react with it
Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
Answer:
Excited state of an electron is the state attained by an electron after it has absorbed energy and it moves further from the nucleus.
an electron is at higher energy when excited and at lower energy when at ground state.
an excited electron is less stable due to the decrease in the nuclear force of attraction and the grounded electron is more stable due to it's close distance to the nucleus.
Phosphorus is in group 15 meaning it have 5 valence electrons. This means that it needs 3 more electrons to create a full outer shell. As these three electrons are negatively charged it means that P is a 3- ion (it’s an anion [negatively charged ion])
Answer:
It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8
Explanation:
The equation that represents a first-order kinetics is:
Ln ([A] / [A]₀] = -kt
<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>
<em />
As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8
Replacing:
Ln ([A] / [A]₀] = -kt
Ln (1/8) = -1.57x10⁷s⁻¹*t
t = 1.32x10⁻⁷s
<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>
