Answer:
51 J
Explanation:
The air inside a bicycle tire pump has 27 joules of heat conducted away. By convention, when heat is released, it takes the negative sign, so Q = -27 J.
77.9 joules of work done are being done on the air inside a bicycle tire pump. By convention, when work is being done on the system, it takes the positive sign, so W = 77.9 J
We can calculate the change in the internal energy (ΔU) using the following expression.
ΔU = Q + W
ΔU = (-27 J) + 77.9 J
ΔU = 51 J
To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a <span>number that represents the
number of units in one mole of any substance. This has the value of 6.022 x
10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:
</span>3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg
Answer:
c
Explanation: correct me if im wrong
Answer:
See the explanation
Explanation:
In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group 
.
In the axial position, we have a more steric hindrance because we have two hydrogens near to the group. If we have <u>more steric hindrance</u> the molecule would be <u>more unstable</u>. In the equatorial positions, we don't <u>any interactions</u> because the group is pointing out. If we don't have <u>any steric hindrance</u> the molecule will be <u>more stable</u>, that's why the molecule will <u>the equatorial position.</u>
See figure 1
I hope it helps!
Answer:
(a) The rate of formation of K2O is 0.12 M/s.
The rate of formation of N2 is also 0.12 M/s
(b) The rate of decomposition of KNO3 is 0.24 M/s
Explanation:
(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.
Rate of formation of O2 is 0.3 M/s
Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s
Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.
Rate of formation of N2 = (2×0.3/5) = 0.12 M/s
(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.
Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s
