Answer:
The energy of photon is 33.28×10⁻¹⁴ J
Explanation:
Given data:
Frequency of photon = 5.02×10²⁰ Hz
Energy of photon = ?
Solution:
E = h.f
h = planc'ks constant = 6.63×10⁻³⁴ Js
by putting values,
E = 6.63×10⁻³⁴ Js × 5.02×10²⁰ Hz
Hz = s⁻¹
E = 33.28×10⁻¹⁴ J
The energy of photon is 33.28×10⁻¹⁴ J.
There are two things that could be done. The first is to increase the mass of the objects. The second is to decrease the distance between the center of masses of the objects.
I think it’s C but I’m not sure
The specific heat of the metal, given the data from the question is 0.60 J/gºC
<h3>Data obtained from the question </h3>
The following data were obtained from the question:
- Mass of metal (M) = 74 g
- Temperature of metal (T) = 94 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 26.5 °C
- Equilibrium temperature (Tₑ) = 32 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of gold can be obtained as follow:
According to the law of conservation of energy, we have:
Heat loss = Heat gain
MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)
74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)
C × 4588 = 2761.44
Divide both side by 4588
C = 2761.44 / 4588
C = 0.60 J/gºC
Thus, the specific heat capacity of the metal is 0.60 J/gºC
Learn more about heat transfer:
brainly.com/question/6363778
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Answer:
No, IR should not soely be used to identify molecules
Explanation:
IR is a method that identifies the functional groups in a molecule by deducing the frequency of stretching and vibration of bonds. Each peculiar type of bond has a frequency for the vibration of each bond represented on the IR spectrum.
However, one method is never enough to identify a compound. A combination of methods must always be used to clear up ambiguities arising from overlapping IR frequencies. Also, interpretation of the nuanced peaks of the fingerprint region in IR spectra is quite challenging and only gives a fair idea of the functional groups present in the compound.
Therefore other methods such as NMR, UV-VISIBLE etc should also be involved in the identification of compounds.
