Answer:
The change in internal energy is - 1.19 kJ
Explanation:
<u>Step 1:</u> Data given
Heat released = 3.5 kJ
Volume calorimeter = 0.200 L
Heat release results in a 7.32 °C
Temperature rise for the next experiment = 2.49 °C
<u>Step 2:</u> Calculate Ccalorimeter
Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C
Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C
<u>Step 3:</u> Calculate energy released
Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ
<u>Step 4:</u> Calculate change in internal energy
ΔU = Q + W W = 0 (no expansion)
Qreac = -Qcal = - 1.19 kJ
ΔU = - 1.19 kJ
The change in internal energy is - 1.19 kJ
One mole of a substance is defined by Avogadro as consisting of 6.022 x 1023 atoms. This is Avogadro's number. To calculate the number of atoms in two moles of sodium, use dimensional analysis. 2.0 moles Na x 6.022⋅1023g1mol=1.20⋅1024 atoms of Na
Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.
In this problem, we have density and we have mass so we can plug into the equation and solve for V.
38.6=270.2/V
<em>*Multiply both sides by V*</em>
38.6V=270.2
<em>*Divide both sides by 38.6*</em>
V=7
The volume of the gold nugget is 7cm3.
Hope this helps!!
Answer:
9 g/ml
Explanation:
5 + 3.3 = 8.5
But you should use sig figs.
5 has no tenths, hundredths, and etc.
Therefore the 8.5 rounds up to 9
So the answer is 9.
Answer:
3,964 years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of the element is 5,730 years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
