<span>This problem uses the
relationship between Ka and the concentrations of the ions. Calculations are as follows:</span>
<span>
</span><span>1.9 x 10-5</span>= x^2 / (0.25 - x)
<span>x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3
</span><span>pH = -log (2.2 x 10^-3)</span> = 2.66
In the compound Co2O3, cobalt has an oxidation number of 3, while oxygen's oxidation number is -2.
An oxidation number is a number assigned to an element in chemical combination that represents the number of electrons lost (or gained, if the number is negative) by an atom of that element in the compound.
Answer:
210 moles
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 25.6 L
Molarity = 8.20 M
Mole of solute =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it can be expressed as:
Molarity = mole /Volume
With the above formula, the mole of solute dissolved can be obtained as follow:
Volume of solution = 25.6 L
Molarity = 8.20 M
Mole of solute =?
Molarity = mole /Volume
8.20 = mole / 25.6
Cross multiply
Mole = 8.20 × 25.6
Mole = 209.92 ≈ 210 moles
Thus, the mole of solute dissolved in the solution is 210 moles
Answer:
The answer to your question is Q = 355.64 J
Explanation:
Data
Heat = Q = ?
Temperature 1 = T1 = 20°C
Temperature 2 = T2 = 37°C
mass = m = 5 g
Specific heat = Cp = 4.184 J/g°C
Formula
Q = mCp(T2 - T1)
-Substitution
Q = (5)(4.184)(37 - 20)
-Simplification
Q = (5)(4.184)(17)
-Result
Q = 355.64 J
Answer:
about 327 g of sulfur
Explanation:
Reaction equations.
2S + 2O2 = 2SO2
2SO2 + O2 = 2SO3
2SO3 + 2H2O = 2H2SO4
1.00 kg H2SO4 (1000 g H2SO4 / 1kg H2SO4) x (1 mole H2SO4 / 98.079 g H2SO4) x
(1 mole S / 1 mole H2SO4) x (32.065 g S / 1 mole S)
= 326.93 g S needed to be burned to produce 1.00 kg of H2SO4
