Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.
Answer:
Half-Life = 18 days
Explanation:
Isotope decay follows the equation:
Ln[A] = -kt + ln[A]₀
<em>Where [A] is amount of isotope after time t,</em>
<em>k is rate constant</em>
<em>[A]₀ is initial amount of isotope.</em>
<em> </em>
If we solve rate constant, we can find half-life by using:
Half-life = ln 2 / Rate constant
Replacing in isotope decay equation:
Ln[1/8] = -k*54 days + ln[1]
-2.07944 = -54k
0.0385days⁻¹ = k
Half-Life = ln 2 / 0.0385days⁻¹
<h3>Half-Life = 18 days</h3>
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Oxidation refers to a change in electrons (In this case, loss of electrons). It has no impact on the number of neutrons, or for that matter on any sub-atomic particles in the nucleus. So the number of neutrons (3) remains the same
Answer:
χH₂ = 0.4946
χN₂ = 0.4130
χAr = 0.0923
Explanation:
The total pressure of the mixture (P) is:
P = pH₂ + pN₂ + pAr
P = 443.0 Torr + 369.9 Torr + 82.7 Torr
P = 895.6 Torr
We can find the mole fraction of each gas (χ) using the following expression.
χi = pi / P
χH₂ = pH₂ / P = 443.0 Torr/895.6 Torr = 0.4946
χN₂ = pN₂ / P = 369.9 Torr/895.6 Torr = 0.4130
χAr = pAr / P = 82.7 Torr/895.6 Torr = 0.0923
Answer:
Explanation:
Hello.
In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:
And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:
Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:
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