The motion causes day and night because the earth is constantly spinning on it's axis.
Answer:
(E) changing temperature
Explanation:
Consider the following reversible balanced reaction:
aA+bB⇋cC+dD
If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:
Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)
where:
[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1
There are some important things to remember when calculating Kc:
- <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes
- Pure solids and liquids, including solvents, are not considered for equilibrium expression.
- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc
After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
More about half-life: brainly.com/question/1160651
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The question is incomplete, the complete question is;
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Answer:
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Explanation:
In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.
The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.
