Answer:
Its D Are in random motion
i hope it will help you if not soo sorry :)
Answer:
Specific gravity of mercury is 13.56 and it is an unit-less quantity.
Explanation:
Mass of the mercury = m = 607.0 lb = 275330.344 g
1 lb = 453.592 g
Volume of the mercury = v = 

Density of the mercury = d=
d = 13.56 g/mL
Specific gravity of substance = Density of substance ÷ Density of water

Specific gravity of mercury :
(unit-less quantity)
Explanation:
<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these cases</u></em>
<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these caseshope it helps you like me plz</u></em>
To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 V1 / P2
V2 = 153 x 3.00 / 203
<span>V2 = 2.26 L</span>
Answer:
The answer to your question is Volume = 11.4 L
Explanation:
Data
Volume 1 = V1 = 6 L
Pressure 1 = P1 = 1 atm
Temperature 1 = T1 = 22°C
Volume 2 = V2 = ?
Pressure 2 = 0.45 atm
Temperature 2 = -21°C
Process
1.- Convert temperature (°C) to °K
T1 = 273 + 22 = 295°K
T2 = 273 + (-21) = 252°K
2.- Use the combined gas law to solve this problem
P1V1 / T1 = P2V2 / T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (6)(1)(252) / (295)(0.45)
- Simplification
V2 = 1512 / 132.75
- Result
V2 = 11.38 L