Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.
In this problem, we have density and we have mass so we can plug into the equation and solve for V.
38.6=270.2/V
<em>*Multiply both sides by V*</em>
38.6V=270.2
<em>*Divide both sides by 38.6*</em>
V=7
The volume of the gold nugget is 7cm3.
Hope this helps!!
<span>C. 11.2 L
There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is
22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match.
Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
Answer:
Explanation:
From the information given:
Mass of carbon tetrachloride = 5 kg
Pressure = 1 bar
The given density for carbon tetrachloride = 1590 kg/m³
The specific heat of carbon tetrachloride = 0.84 kJ/kg K
From the composition, the initial volume of carbon tetrachloride will be:
= 0.0031 m³
Suppose
is independent of temperature while pressure is constant;
Then:
The change in volume can be expressed as:





However; the workdone = -PdV

W = - 7.6 J
The heat energy Q = Δ h


Q = 84 kJ
The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;
ΔU = ΔQ + W
ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)
ΔU = 83.992 kJ
Answer:
W = -120 KJ
Explanation:
Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.
Thus; T1 = T2 = 400K
change in entropy; ΔS = −0.3 kJ/K
Formula for change in entropy is written as;
ΔS = Q/T
Where Q is amount of heat transferred.
Thus;
Q = ΔS × T
Q = -0.3 × 400
Q = -120 KJ
From the first law of thermodynamics, we can find the workdone from;
Q = ΔU + W
Where;
ΔU is Change in the internal energy
W = Work done
Now, since it's an ideal gas model, the change in internal energy is expressed as;
ΔU = m•C_v•ΔT
Where;
m is mass
C_v is heat capacity at constant volume
ΔT is change in temperature
Now, since it's an isothermal process where temperature is constant, then;
ΔT = T2 - T1 = 0
Thus;
ΔU = m•C_v•ΔT = 0
ΔU = 0
From earlier;
Q = ΔU + W
Thus;
-120 = 0+ W
W = -120 KJ