Answer:
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Explanation:
Molecular Equation:
Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)
Ionic Equation:
Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)
Net Ionic Equation: => Drop NO₃⁻ as spectator ion
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Answer:
the answer for that is false
Answer:
Explanation:
wavelength λ = 12.4 x 10⁻² m .
energy of one photon = h c / λ
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 12.4 x 10⁻²
= 1.6 x 10⁻²⁴ J .
Let density of coffee be equal to density of water .
mass of coffee = 255 x 1 = 255 g
heat required to heat up coffee = mass x specific heat x rise in temp
= 255 x 4.18 x ( 62-25 )
= 39438.3 J .
No of photons required = heat energy required / energy of one photon
= 39438.3 / 1.6 x 10⁻²⁴
= 24649 x 10²⁴
= 24.65 x 10²⁷ .
Answer:
She should not have multiplied the nitrogen atom by subscript 2.
Explanation:
Chemical formula:
3(NH₄)₂SO₄
Elements present in given formula:
Nitrogen
Hydrogen
Sulfur
Oxygen
Total number of atoms of elements:
N = 3×1×2 = 6
H = 4×2×3 = 24
S = 1×3 = 3
O = 3×4 = 12
The number nitrogen atoms are six. Elena did mistake by counting the number of nitrogen. She should didn't multiplied the nitrogen atom by subscript 2.
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.
