Actually the correct answer must be:
The limiting reactant in the reaction is the one which has
the lowest ratio of moles available
over coefficient in the balanced equation
This is because the actual mass or number of moles of the
reactant does not directly dictate if it is a limiting reactant, this must be
relative to the other reactants.
So the answer is:
e. none of the above
The overall equation for the combined reactions become:
2VO₂⁺²₍aq₎ + 8H⁺₍aq₎ + 3Zn⁺²⁽s⁾ ⇒ 2V⁺²₍aq₎ + 3Zn⁺²₍aq₎ + 4H₂O₍l₎
The volume of solution is:
11.7/1000 = 0.0117 Litres
The moles of VO₂⁺² are:
0.0117 × 0.0037 = 4.3 × 10⁻⁵ mol
As per the equation, 2 moles of VO₂⁺² need 3 moles of 3Zn⁺²
Therefore, moles of Zn⁺² needed are:
6.5 × 10⁻⁵ mol
One mole Zn metal produces one mol of ions. So we need
6.5 × 10⁻⁵ mol of Zn metal
Mass required = moles × Molecular weight
Mass = 6.5 × 10⁻⁵ × 65
Mass = 0.0042 grams
Answer:
<h3>The answer is 0.75 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 90 g
volume = 120 mL
We have
We have the final answer as
<h3>0.75 g/mL</h3>
Hope this helps you
The ionization energy is 1 × 10^-18 J
<h3>What is ionization energy?</h3>
The ionization energy is the energy required to remove an electron from an atom.
We have the following information;
wavelength of the photon = 58.4 nm
Speed of the electron = 2310 × 10^3 m/s
Since;
hv = I + 1/2mv^2
v = c/λ
hc/λ = I + 1/2mv^2
I = hc/λ - 1/2mv^2
I = (6.6 × 10^-34 × 3 × 10^8/58.4 × 10^-9) - (1/2 × 9.11 × 10^-31 × (2310 × 10^3)^2)
I = 1 × 10^-18 J
Learn more about ionization energy: brainly.com/question/16243729
Let's assume that CH₄ has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT
Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
P = 1 atm = 101325 Pa
V = 1.50 L = 1.50 x 10⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
T = 0 °C = 273 K
By substitution,
101325 Pa x 1.50 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K
n = 0.0669 mol
Hence, moles of CH₄ = 0.0669 mol
Moles = mass / molar mass
Molar mass of CH₄ = 16 g mol⁻¹
Mass of CH₄ = moles x molar mass
= 0.0669 mol x 16 g mol⁻¹
= 1.0704 g
Hence, mass of CH₄ in 1.50 L at STP is 1.0704 g