Answer:
Explanation:
The <em>purchase price</em> is what Janice invested for every share.
Since the stock was priced at $31.82 per share and she received a $1.11 dividend per share, her investment was:
- $31.82 - $1.11 = $30.71 per share ← answer
This price is the cost for Janice, over which she shall calculate their returns (gains or losses) on the future, when she sells the shares, for instance.
The total investment of Janice was the number of shares multipled by the purchase price:
- 40 shares × ($31.82 - $1.11)/ share
- 40 shares × ($30.71) / share = $1,228.40 (total investment)
Answer:
A. The pressure will increase 4 times. P₂ = 4 P₁
B. The pressure will decrease to half its value. P₂ = 0.5 P₁
C. The pressure will decrease to half its value. P₂ = 0.5 P₁
Explanation:
Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.
<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>
<em />
<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>
V₂ = 0.25 V₁. According to Boyle's law,
P₁ . V₁ = P₂ . V₂
P₁ . V₁ = P₂ . 0.25 V₁
P₁ = P₂ . 0.25
P₂ = 4 P₁
<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>
T₂ = 0.5 T₁. According to Gay-Lussac's law,

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>
n₂ = 0.5 n₁.
P₁ in terms of the ideal gas equation is:

P₂ in terms of the ideal gas equation is:

To increase the energy of the emitted electrons, the frequency of the incident light on the metal must be increased.
<h3>What is energy of emitted electron?</h3>
The maximum energy of an emitted electron is equal to the energy of a photon for frequency f (E = hf ), minus the energy required to eject an electron from the metal's surface, also known as work function.
Ee = E - W
<h3>Energy of the emitted electron</h3>
The energy of emitted electrons based on the research of Albert Einstein is given as;
E = hf
where;
- h is planck's constant
- f is frequency of incident light on the metal
Thus, to increase the energy of the emitted electrons, the frequency of the incident light on the metal must be increased.
Learn more about energy of electron here: brainly.com/question/11316046
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Answer:
here you go
Explanation:
Halogens are very electronegative. This means that inductively they are electron withdrawing. However, because of their ability to donate a lone pair of electrons in resonance forms, they are activators and ortho/para directing. Electron withdrawing groups are meta directors and they are deactivators.