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JulsSmile [24]
2 years ago
14

eleanor purchased $2568 worth of stock and paid her broker a 0.5% fee. She sold the stock when the stock price increased to 3928

using an online broker that charged $7 per trade. What are her net proceeds?
Chemistry
1 answer:
Mrrafil [7]2 years ago
5 0

Answer: $1338.16

Explanation: Total cost of stock= $2568

Total cost of stock including the brokerage =2568+\frac{0.5}{100}\times {2568}=2580.84$

Selling price of stock = $3928

Selling price of stock including trading fee=($3928-$7)=$3919

Net Proceeds = Net selling price of stock - Cost Price of stock

Net Proceeds = ($3919 - $2580.84) = $1338.16


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Study the following unbalanced half-reaction. Which equation represents the balanced half-reaction? H2O2 ---> H2O
Aleonysh [2.5K]
O: 1*2 = 2*1 
<span>H 2 + 2 = 2*2 </span>

<span>answer C hope you get it right</span>
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2 years ago
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Which element in period 4 would have chemical properties similar to magnesium? 7. which metalloids would have chemical propertie
azamat
The only solution you'll need to have for this problem is a periodic table. The columns in the table are called groups, and they are number from the left to the right starting with 1. The rows in the table are called periods which are numbered from the top to bottom starting with 1. 

6. Elements that belong to the same group portray similar chemical properties. Therefore, the element in period 4 which is also in group 2 is <em>Calcium (Ca)</em>.

7. The elements that are striked through with the red slanting lines are the metalloids. All the elements to the left of the metalloids are metals. All the elements to the right are nonmetals. Bromine has a symbol of Br. Since At is a metalloid and located in the same group with Br, the <em>answer is Astatine (At).</em>

8. Tin has the chemical symbol of Sn. The nonmetal that is located in the same group is <em>Carbon (C)</em>.

9. All the elements in period 6 would have similar properties. The answers could be: <em>Phosphorus (P), Arsenic (As), Antimony (Sb) and Bismuth (Bi)</em>.

10. Period is row 1 and group 18 is the last column. <em>The answer is Helium (He).</em>

5 0
2 years ago
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(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

6 0
2 years ago
Caleb wants to find out which type of floor material is best for bouncing tennis balls the highest. Which statement best describ
goldenfox [79]
Option 2 seems to be the best answer
4 0
2 years ago
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Submit At 25.0 C, a 10.00 L vessel is filled with 5.00 atm of Gas A and 7.89 atm of Gas B. What is the mole fraction of Gas B?
Firlakuza [10]

Answer:

the mole fraction of Gas B is xB= 0.612 (61.2%)

Explanation:

Assuming ideal gas behaviour of A and B, then

pA*V=nA*R*T

pB*V=nB*R*T

where

V= volume = 10 L

T= temperature= 25°C= 298 K

pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm

R= ideal gas constant = 0.082 atm*L/(mol*K)

therefore

nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole

nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole

therefore the total number of moles is

n = nA +nB= 2.04 mole +  3.22 mole = 5.26 mole

the mole fraction of Gas B is then

xB= nB/n= 3.22 mole/5.26 mole = 0.612

xB= 0.612

Note

another way to obtain it is through Dalton's law

P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612

5 0
2 years ago
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