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Scorpion4ik [409]
3 years ago
9

Ava, Inc., issued 7% bonds, dated January 1, with a face amount of $117,200 on January 1, 2016 for an issue price of 106.5. The

bonds mature on December 31, 2025 (10 years). For bonds of similar risk and maturity the market yield is 9%. Interest is paid annually on December 31.
What is the interest payment on the bond?

10b) A company issued 5%, 20-year bonds with a face amount of $302,000. The market yield for bonds of similar risk and maturity is 5%. Interest is paid annually. What is the debit to cash for the bond proceeds?

Here are the PV and PVA factors

PV i=5%, 20 periods

PVA i=5%, 20 periods

PV i=7%, 20 periods

PVA i=7%, 20 periods

.37689

12.46221

.25842

10.59401

Chemistry
1 answer:
natta225 [31]3 years ago
8 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

10a The interest payment on the bond is = $ 8,204

10b The debit to cash for the bond proceeds is = $ 302,000

Explanation:

The explanation is shown on the second uploaded image

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Hydrofluoric acid (HF) can be prepared according to the following equation:
Artemon [7]

Given the balanced equation:

( Reaction type : double replacement)

CaF2 + H2SO4 → CaSO4 + 2HFI

We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.

This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.

For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.

For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.

Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.

3 0
3 years ago
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What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?
Vanyuwa [196]

Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

Explanation:

22.3% by mass → In 100 g of solution, we have  22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g.  1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

6 0
3 years ago
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He orbital period of an object is 2 × 107 s and its total radius is 4 × 1010 m
gtnhenbr [62]

Answer:

The answer is 12,560  

Explanation:

The orbital period is the time a given cosmic question takes to finish one circle around another protest and applies in space science as a rule to planets or space rocks circling the Sun, moons circling planets, exoplanets circling different stars, or double stars. Mercury, for instance, has an orbital time of 88 days while it takes Jupiter around 11.86 years. The time of the Earth's circle is generally thought to be 365 days as timetables appear.

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EFFECTS Of The<br> HUMAN INTERVATION OF THE<br> CARBON CYCLE.
vladimir2022 [97]

Human activities have a tremendous impact on the carbon cycle. Burning fossil fuels, changing land use, and using limestone to make concrete all transfer significant quantities of carbon into the atmosphere. ... This extra carbon dioxide is lowering the ocean's pH, through a process called ocean acidification.

8 0
2 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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