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frez [133]
3 years ago
5

Which of the following is true about a municipal bond with a put option? (A) An investor will exercise the option to put the bon

d if yields rise significantly. (B) An investor must have the issuer's permission to put the bond. (C) The market price of bonds is never affected by a put option feature. (D) Yields are usually higher for a new issue bond with a put option than for a new issue bond without a put option.
Chemistry
1 answer:
vampirchik [111]3 years ago
5 0

Answer:

(A) An investor will exercise the option to put the bond if yields rise significantly

Explanation:

A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.

Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.

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d. Butanol has the chemical formula C4H,OH. Write and balance the chemical reaction for the combustion of butanol.​
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Is hardness physical or chemical property
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Hydrazine (N 2 H 4 )) a rocket fuel reacts with oxygen to form nitrogen gas and water vapor . The reaction is represented with t
Drupady [299]

The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)

<h3>Balanced equation </h3>

N₂H₄ + O₂ —> N₂ + 2H₂O

Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol

Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g

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Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

<h3>How to determine the mass of N₂H₄</h3>

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

Therefore,

96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄

Thus, 85.4 g of N₂H₄ is needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

8 0
3 years ago
30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?
USPshnik [31]
Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 &deg; C = 23+273.15 = 296.15 &deg; K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

</span>
5 0
3 years ago
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