Answer:
6e⁻
Explanation:
Charge in reducing 2 mole of Cr⁶⁺ to Cr³⁺
Cr⁶⁺ + xe⁻ → Cr³⁺
The numbers 6+ and 3+ on the chromium atoms are the oxidation number of the atom.
It shows the number of electrons that has been lost by chromium in an oxidation state.
Cr⁶⁺ shows that it has lost 6 electrons. When an atom loses electrons, the number of protons becomes more. This makes it positively charged in nature.
This analogy goes for Cr³⁺
Therefore, in going from reactants to products, chromium
6+ → 3+
In this kind of expression, the number of atoms must be conserved and the charges too;
2Cr⁶⁺ + xe⁻ → 2Cr³⁺
given 2 moles of Cr;
To balance the charge;
2(6+) + x(-) = 2(3+)
12 - x = 6
x = 6
A charge of 6e⁻ is required to reduce 2 mole of Cr⁶⁺ to Cr³⁺
Answer:
1900 °C
Step-by-step explanation:
This looks like a case where we can use the <em>Combined Gas Law</em> to calculate the temperature.
p₁V₁/T₁ = p₂V₂/T₂ Multiply both sides by T₂
p₁V₁T₂/T₁ = p₂V₂ Multiply each side by T₁
p₁V₁T₂ = p₂V₂T₁ Divide each side by p₁V₁
T₂ = T₁ × p₂/p₁ × V₂/V₁
=====
Data:
We must convert the pressures to a common unit. I have chosen atmospheres.
p₁ = 675 mmHg × 1atm/760 mmHg = 0.8882 atm
V₁ = 718 mL = 0.718 L
T₁ = 48 °C = 321.15 K
p₂ = 159 kPa × 1 atm/101.325 kPa = 1.569 atm
V₂ = 2.0 L
T₂ = ?
=====
Calculation:
T₂ = 321.15 × 1.569/0.8882 × 2.0/0.718
T₂ = 321.15 × 1.766 × 2.786
T₂ = 321.15 × 1.569/0.8882 × 7.786
T₂ = 1580K
T₂ = 1580 + 273.15
T₂ = 1900 °C
<em>Note</em>: The answer can have only <em>two</em> significant figures because that is all you gave for the second volume of the gas.
Because they have different boiling points
Answer:
0.508 L of solution.
Explanation:
Always a safe bet to convert to moles:

Where n is moles, m is mass, and MM is molar mass.
Now remember the equation for concentration (molarity):

Where C is the concentration, n is moles, and V is volume.
To make this easy, combine the two equations (note n appears in both, so you can do a substitution) and solve for V as the question asks:

Therefore we can make 0.508 L of solution.