What is meant by the statement "energy is quantized"?
1 answer:
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Answer: The density of the material is 2.66 g/mL and it is likely this is made of Aluminum
Explanation:
The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = . Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:
This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.
has 37 electrons in the orbit about its nucleus.
Atomic mass is the total number of neutrons and protons present in the nucleus of an element.
The atomic number is the total number of protons present in the nucleus of an element.
For writing an atomic symbol, atomic mass is written on the top left of the atomic symbol, and the atomic number is written on the downside of the left of the symbol.
An electrically neutral atom is an atom that has an equal number of electrons as well as protons.
Thus from the above conclusion we can say that has 37 electrons.
Learn more about Atomic mass and atomic number here:
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Answer:
a. 478.69 K
b. 939.43
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
= 785
= 0.000785
= 400K
= 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 ⋅Pa⋅
⋅
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5 - (34980 + 35.5
)]
83.8 = (0.03 × 35.5) ( - 400K)
83.8 = 1.065 ( - 400)
78.69 = ( - 400)
= 400 + 78.69
= 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
/
=
/
=
/
= (785 × 478.69)/400
= 939.43
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P(-
) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J