I had this same question a while back, so I may not be right, but I'm pretty sure it's D. None of the above.
<span>We can use the ideal gas law PV=nRT
For the first phase
The starting temperature (T1) is 273.15K (0C). n is 1 mole, R is a constant, P = 1 atm, V1 is unknown.
The end temperature (T2) is unknown, n= 1 mol, R is a constant, P = 3*P1= 3 atm, V2=V1
Since n, R, and V will be constant between the two conditions: P1/T1=P2/T2
or T2= (P2*T1)/(P1) so T2= (3 atm*273.15K)/(1 atm)= 3*273.15= 816.45K
For the second phase:
Only the temperature and volume change while n, P, and R are constant between the start and finish.
So: V1/T1=V2/T2 While we don't know the initial volume, we know that V2=2*V1 and T1=816.45K
So T2=(V2*T1)/V1= (2*V1*T1)/V1=2*T1= 2*816.45K= 1638.9K
To find the total heat added to the gas you need to subtract the original amount of heat so
1638.9K-273.15K= 1365.75K</span>
Explanation:
Sublimation describes a solid turning directly into a gas. Melting, on the other hand, occurs when a solid turns into a liquid. Water can, under the right circumstances, sublimate, though it usually melts at temperatures above 0 degrees Celsius or 32 degrees Fahrenheit. Carbon dioxide (CO2), however, is very different. The conditions that determine whether CO2 melts or sublimates are both temperature and atmospheric pressure.
The unknown alcohol = 2-propanol
<h3>Further explanation</h3>
Given
volume = 50 ml
mass = 39.4 g
Required
The unknown alcohol
Solution
Density of sample :
= mass : volume
= 39.4 g : 50 ml
= 0.788 g/ml
The unknown alcohol at room temperature = liquid, so the unknown alcohol is 2-propanol because it has temperature range between -90 to 82 for liquid and the density = 0.788 g/ml
Explanation:
The four quantum numbers gives the position of the orbital, its spatial orientation, the shape of the orbital and the spin of the an electron in the orbital.
The principal quantum number(n) gives the main energy level in which the orbital is located; n = 1,2,3,4....
Azimuthal or secondary quantum number (l) gives the shape of the orbitals an they take values l = 0,1,2,3.... and their names are s for 0, p for 1, d for 2 and f for 3.
Magnetic quantum number gives the spatial orientation or degeneracy of the orbitals in the subshell.
Spin quantum number gives the spinning of the electrons
Electronic configuration:
Na = 1s² 2s² 2p⁶ 3s¹
The last electron enters the 3s¹
n = 3
l = 0
m = 0
spin =+ or - 
N = 1s² 2s² 2p³
last electron enters 2p³
n = 2
l = 1
m = -1, 0, 1
spin = + or -
O = 1s² 2s² 2p²
last electron enters 2p²
n = 2
l = 1
m = -1, 0, 1
spin = + or -
Cl =1s² 2s² 2p⁶ 3s² 3p⁵
last electron enters 3p⁵
n = 3
l = 1
m = -1, 0, 1
spin = + or -
S = 1s² 2s² 2p⁶ 3s² 3p⁴
last electron enters 3p⁴
n = 3
l = 1
m = -1, 0, 1
spin = + or -
learn more:
Quantum number brainly.com/question/9288609
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