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JulsSmile [24]
1 year 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]1 year 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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If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is____J.
astra-53 [7]

Answer:If a bouncing ball has a total energy of 20 J and a kinetic energy of 5 J, the ball’s potential energy is 15J.

If the kinetic energy of the ball decreases, then the potential energy will Increase.

Explanation:

5 0
1 year ago
Based on the activity series provided, which reactants will form products?
allsm [11]

Answer: CuI₂ + Br₂

Explanation:

1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).

The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.

2) Choice 1: CuI₂ + Br₂

Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:

CuI₂ + Br₂ → CuBr₂ + I₂

Being I less active than Br, it cannot displace Br in CuBr₂.

3) Choice 2: Cl₂ + AlF₃

Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.

4) Choice 3: Br₂ + NaCl

Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.

5) Choice 4: CuF₂ + I₂

Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.

6 0
2 years ago
Read 2 more answers
A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa
bogdanovich [222]

Explanation:

The given data is as follows.

         V_{1} = 85.0 ml,        M_{1} = 0.9 M

         V_{2} = 85.0 ml,        M_{1} = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = 0.9 M \times 0.085 L        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = mC \Delta T

                     4271 J = 170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C    

                     0.0773 = T_{f} - 18.24

                    T_{f} = 18.317^{o}C  

Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

6 0
2 years ago
Part B: Copper (II) chloride (CuCl2; 0.98g) was dissolved in water and a piece of aluminum wire (Al; 0.56g) was placed in the so
babymother [125]

.,. .................

7 0
1 year ago
Consider the electrolysis of aqueous agno3. Refer to the table of standard reduction potentials as needed. What should form at t
Valentin [98]

Answer:

Ag+

Explanation:

anode: 2AgNO3(l)⟶Ag(aq)+NO3(g)

8 0
1 year ago
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