The mass of a sample of alcohol is found to be = m = 367 g
Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.
Explanation:
The Energy of the sample given is q = 4780
We are required to find the mass of alcohol m = ?
Given that,
The specific heat given is represented by = c = 2.4 J/gC
The temperature given is ΔT = 5.43° C
The mass of sample of alcohol can be found as follows,
The formula is c = ![\frac{q}{mt}](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7Bmt%7D)
We can drive value of m bu shifting m on the left hand side,
m = ![\frac{q}{ct}](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7Bct%7D)
mass of alcohol (m) = ![\frac{4780}{(2.4)( 5.43)}](https://tex.z-dn.net/?f=%5Cfrac%7B4780%7D%7B%282.4%29%28%205.43%29%7D)
m = 367 g
Therefore, The mass of the given sample of alcohol is
m = 367g
It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.
Answer:
5: 0.16
6: 50
Explanation:
Question 5:
We can use the equation density = mass/ volume.
We already have the mass (12g), but now we need to find the volume of the cylinder.
The equation for this is πr²h
So we know the radius is 2 and the height is 6.
π x (2)² x 6 = 24π = 75.398cm³
Now we can use the density equation above:
12/75.398 = 0.1592g/cm³ = 0.16g/cm³.
Question 6:
This time, we have to rearrange the equation density = mass/ volume to find the mass.
We know mass = density x volume.
From the question, the density is 2.5g/mL and the volume is 20mL.
Following the equation above, we do 2.5 x 20 to get 50g.
Answer:
6.02×10^23 atoms
Explanation:
Avogadros constant is a number that states the amount of atoms in one mole of a substance which is 6.02×10^23 to 3 significant figures.
Answer:
En general, la adicción se considera una enfermedad crónica que ocurre tras el consumo continuado de una droga durante un período de tiempo relativamente largo. Por tanto, el principal factor desencadenante de la enfermedad es el consumo mismo de la droga.Sep 16, 2019
Explanation: