Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13
Answer:
1.84 L
Explanation:
Using the equation for reversible work:

Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*(
- 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,

When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation,
= 102 pm
Radius of cation,
= 196 pm

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
To answer this, we look at the polarities of these molecules given. A molecule is said to be polar when there is an unequal shraing of electrons the opposite is called nonpolar.
CH4 = nonpolar
CH3OH = polar
CH3 CH3 = nonpolar
CH3 CH2 CH2 OH = polar
<span>CH3 CH2 CH2 CH2 CH2 CH3 = nonpolar</span>
Answer:
3,46 atm
Explanation:
Tenemos;
P1 =?
T1 = 100 + 273 = 373 K
V1 = 40 litros
P2 = 6 atmósferas
T2 = 50 + 273 = 323 K
V2 = 20 litros
Desde;
P1V1 / T1 = P2V2 / T2
P1V1T2 = P2V2T1
P1 = P2V2T1 / V1T2
P1 = 6 * 20 * 373/40 * 323
P1 = 44760/12920
P1 = 3,46 atm