How do you calculate temperature?
Does anyone understand task 3? If so, please provide the calculation method. I'm at a loss.
Does anyone understand task 3? If so, please provide the calculation method. I'm at a loss.
Hello, I'm currently practicing for my chemistry assignment and have a question about one of the tasks. If lead oxide (PbO2) is heated on charcoal, lead is formed. I would be very happy to receive an answer with an explanation. Thank you in advance.
Hello, here is the picture for context: My solution would be the following: I substitute the values p/R for Q. Then I take the inverse of the equation and cancel p. Now r is in the denominator on the right side and R is in the numerator on the right. 1 to the power of…
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Hello dear community, I have the following function: And the derivations are: Now a Taylor polynomial of degree 2 is to be set up for fn(x) and then a remainder term is to be estimated from the third derivative, which looks like this: And I don't understand this estimate. Where does sqrt(n)/2 come from? Can…
For task 3:
First determine the temperature difference between 18° C. and -35° C. This is called ΔT (“Delta T”). ΔT has a negative amount in this case because the temperature decreases.
Now you need to paint ΔT with the volume expansion coefficient γ (“Gamma”) of brass. This results in the relative volume change ΔV/V around which the brass is contracted when it is cooled by ΔT. Attention, the unit of γ is incorrectly specified in the task: it is not cm3/°C, but 1/°C. If you calculate ΔT times γ, then °C is abbreviated and the result, i.e. ΔV/V, is a number without unit, in this case a negative number.
Now you can consider what the new volume V2 of the brass is. Before it was V, and now it is V+ΔV. With the relative volume change ΔV/V, you can then say: V2 =V*(1+ΔV/V). It follows that the volume has changed by the factor V2/V=1+ΔV/V.
Now we come to the solution, the changed density. The density = mass/volume. The mass has remained the same. The new density ρ2 is therefore mass/new volume.
ρ = m/V
ρ2 = m/V2
The ratio of the new to the old density is thus:
ρ2/ρ = V/V2
ρ2/ρ = 1/(1+ΔV/V)
ρ2 = ρ/(1+ΔV/V)
Since we have cooled the brass and ΔV is negative, 1+ΔV/V < 1, i.e. the density has increased, i.e. ρ2 is greater than ρ.
Task 3:
it cools down by 53°C (from +18 to -35).
Calculate the material.
So 53°C * 0,000 057 cm3/°C = 0.003 cm3 (As much shrinks 1 cm3)
Before 1cm3 weighs exactly 8.1 g
Now the one cm3 has shrunk to 1 – 0.003 = 0.997 cm3, but weighs as much as before.
Now only, convert again to 1 cm3 by 3-set for the new density.
8.1 g/cm3 / 0.997cm3 * 1cm3 = 8.124 g/cm3
C) is dipped. Iron pulls together with cold, so it is shorter and gives a higher value