The correct option is (C) 6.02 X 10²³
A sample of CH₄O with a mass of 32.0 g contains <u>6.02 X 10²³</u> molecules of CH₄O.
To calculate the number of moles;
Molar mass of CH₄O = C + 4(H) + O
= 12.01 + 4(1.008) + 16
= 32.04 g/mol
So, 1 mol of CH₄O = 32.04 g of CH₄O
Given, 32.0 g of CH₄O
According to Avagadro's constant 1 mole of a substance contains 6.022× 10^23 particles (molecules, atoms or ions).
= (32.0 g/1)(1 mol CH₄O/32.04 g CH₄O)(6.02x10²³/1 mol CH₄O)
= 6.02 X 10²³ molecules of CH₄O
Hence, a sample of will contain number of molecules 6.02 X 10²³ molecules.
Learn more about the Moles calculation with the help of the given link:
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We can use the magnification equation for this problem.
Magnification equation:
<em>hi/h0 = di/d0</em>
Where in <em>hi </em>: image size, <em>h0</em> : object size, <em>di </em>: image distance, <em>d0 </em>: object distance from mirror
So plugging in the given variables we will have the corresponding equation
0.1 m / 0.3 m = <em>di / </em>0.4 m
<em>di</em> = 0.1333333m
The generated image of the object is located 0.13 meters or 13 cm away from the mirror
The initial temperature is 137.34 °C.
<u>Explanation:</u>
As the specific heat formula says that the heat energy required is directly proportional to the mass and change in temperature of the system.
Q = mcΔT
So, here the mass m is given as 23 kg, the specific heat of steel is given as c = 490 J/kg°C and the initial temperature is required to find with the final temperature being 140 °C. Also the heat energy required is 30,000 J.
ΔT =
ΔT =
Since the difference in temperature is 2.66, then the initial temperature will be
Final temperature - Initial temperature = Change in temperature
140-Initial temperature = 2.66
Initial temperature = 140-2.66 = 137.34 °C
Thus, the initial temperature is 137.34 °C.
Answer:
In a compound, chemical energy is _____required__________ when bonds break.
Answer:
25.97oC
Explanation:
Heat lost by aluminum = heat gained by water
M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]
Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC
Let Temp(Al+H2O) = X
23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)
21.15(65.9-X) = 230.23(X-22.3)
1393.785 - 21.15X = 230.23X – 5134.129
230.23X + 21.15X = 1393.785 + 5134.129
251.38X = 6527.909
X = 6527.909/251.38
X = 25.97oC
So, the final temperature of the water and aluminum is = 25.97oC
