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erastovalidia [21]
3 years ago
13

One year ago, you sold a put option on 100,000 euros with an expiration date of 1 year. you received a premium on the put option

of $0.04 per unit. the exercise price was $1.22. assume that 1 year ago the spot rate of the euro was $1.20, the 1-year forward rate exhibited a discount of 2 percent, and the 1-year futures price was the same as the 1-year forward rate. from 1 year ago to today, the euro depreciated against the dollar by 4 percent. today the put option will be exercised (if it is feasible for the buyer to do so).
Chemistry
1 answer:
pochemuha3 years ago
5 0
There are two problems for this question:1. What is the total dollar amount of your profit and loss:
Put option premium is equal to 0.04 per unit.
The exercise price is 1.22
One option contract is 100,000
Selling price is 1.20
-Purchase prise is - 1.22
-Premium paid is +0.04
Net profit is = 0.02 x 100,000 = 2,000 – 80 = 1,920

2. Now undertake that as an alternative of taking a position in the put option one year ago, you sold a future's contract on 100,000 euros with a payment date of one year.
Find the total dollar amount of your profit or loss.
Solution: Contract to buy: $1.20 x 100,000 = 120,000 at payment date.
Contract to sell: $1.22 x 100,000 = 122,000 at settlement date
Settle contracts: -2,000 - 80 = -$2,080
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8 0
3 years ago
PLEASE HELP
blondinia [14]

Answer:

c elevation

Explanation:

7 0
2 years ago
Read 2 more answers
What is the new boiling point if 25 g of NaCl is dissolved in 1.0 kg of water
Inessa05 [86]
The correct answer for the question that is being presented above is this one: 

Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m 
<span>delta Tf = Kfm Kf H2O = 1.86 degrees C/m 
</span>
We need to know the formula for Molality.
molality = mol solute / kg solvent 

<span>We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. </span>

<span>25 g NaCl / 58.5 g/mol = 0.427 mol </span>

<span>Then, use the formula for molality. </span>

<span>molality = mol solute / kg solvent </span>
<span>= 0.427 / 1 </span>
<span>= 0.427 m </span>

<span>Use now the formula to get the boiling point.</span>

<span>delta Tb = Kbm </span>
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8 0
3 years ago
How much energy is required to raise the temperature of a 300.0
Montano1993 [528]

Answer:

Q = 1455.12 Joules.

Explanation:

Given the following data;

Mass = 300 grams

Initial temperature = 22.3

Final temperature = 59.9°C

Specific heat capacity = 0.129 J/gºC.

To find the quantity of energy;

Q = mcdt

Where,

Q represents the heat capacity.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt represents the change in temperature.

dt = T2 - T1

dt = 59.9 - 22.3

dt = 37.6°C

Substituting the values into the equation, we have;

Q = 300*0.129*37.6

Q = 1455.12 Joules.

8 0
3 years ago
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

7 0
2 years ago
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