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Inessa05 [86]
3 years ago
9

what's the main difference between continuous compounding and exponential growth, and why does my text book say that it is bette

r to express growth rates as if they are continuously compounded? What's the advantage---both give outrageous estimates to "what if" scenarios. ...?
Chemistry
1 answer:
Annette [7]3 years ago
4 0
The big advantage to using continuous compounding to express growth rates is it avoids the problem of asymmetry in growth rates:

For example, if we use the normal definition and $100 grows to $105 in one time period, that's a growth rate of $105/$100 - 1 = 5% But if $105 decreases to $100, that's a growth rate of $100/$105 - 1 = -4.76%

The problem of asymmetry is those two growth rates, 5% and -4.75% are not equal up to a sign.

But if you use continuous compounding the growth rate in the first case is ln(105/100) = 0.04879.
And the growth rate in the second is ln (100/105) = -0.04879.
Those two growth rates are definitely the negative of each other.<span>
</span>
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A compound is 52.14% C, 13.13% H, and 34.73% O by mass. What is the empirical formula of the compound?
yarga [219]
Give the compound a mass of 100 grams, so that the mass of each element is
C=52.14% of 100g = 52.14g
H=13.13g
O=34.73g

convert the mass to moles
divide the mole value by the smallest number of moles calculated
round to nearest whole number, this is the mole ratio / subscripts
7 0
3 years ago
Read 2 more answers
Consider the following list of liquids and their boiling points: ether, bp = 35°C acetone, bp = 56°C cyclohexane, bp = 80°C wate
SVEN [57.7K]

Answer:

Ether

Explanation:

NB: The boiling point of the liquids given are referred to as their normal boiling point i.e the boiling point at atmospheric pressure.

The relationship between normal Boiling point and vapor pressure is that; at a given temperature the lower the normal boiling point, the higher the vapor pressure.

hence ether with the least normal boiling point of 35°C is expected to have the largest vapor pressure near room temperature.

8 0
3 years ago
Example 1: urea, (nh2)2co, is used in the manufacture of resins and glues. when 5.00 g of urea is dissolved in 250.0 ml of water
katen-ka-za [31]
A) i believe the reaction is exothermic, because 27.6 kg of thermal energy is gained by the water solution, the dissolution of urea is exothermic. Exothermic reaction is a chemical reaction that releases energy by light or heat. It is the opposite of an endothermic reaction where heat is gained by the reaction.

b) The water gained the heat released when urea dissolve. That is the water gained 27.6 kJ, while dissolution of urea released 27.6 kJ. Therefore, the heat gained is equal to the heat lost.

c) From part B, since heat gained is equal to heal lost, then
250 g × (Tf -30) ×4.18 J/g = 27600 J
 = 1045 Tf - 31350J = 27600 J
Tf = 56.41°C.
Therefore the final temperature  of the solution is 56.41°C

d) The initial and final temperature in Fahrenheit
°F = °C × (9/5) +32,
Thus, 30°C will be equal to 86° F
while 56.41 will be equivalent to  133.54 ° F


7 0
3 years ago
Read 2 more answers
Give the percent yield when 162.8 g of CO2 are formed from the reaction of excess amount of
Zepler [3.9K]

Answer:

84.86%

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction between C8H18 and O2 to produce CO2. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Step 2:

Now, let us calculate the mass of O2 that reacted and the mass of CO2 produced from the balanced equation above. This is illustrated below:

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 25 x 32 = 800g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Therefore the mass of O2 that reacted from the balanced equation is 800g

The mass of CO2 produced from the balanced equation is 704g

Step 3:

Determination of the theoretical yield of CO2. This is illustrated below:

From the balanced equation above,

800g of O2 reacted to produced 704g of CO2.

Therefore, 218g of O2 will react to produce = (218 x 704)/800 = 191.84g of CO2.

Therefore, the theoretical yield of CO2 is 191.84g

Step 4:

Determination of the percentage yield of CO2. This is illustrated below:

Actual yield = 162.8g

Theoretical yield = 191.84g

Percentage yield =?

Percentage yield = Actual yieldm/Theoretical yield x100

Percentage yield = 162.8/191.84 x100

Percentage yield = 84.86%

Therefore, the percentage yield of CO2 is 84.86%

6 0
3 years ago
Read 2 more answers
Why are being flammable and combustible considered to be chemical properties and not physical properties?
g100num [7]

Answer:

C. They change the substance into another substance.

Explanation:

Hope it helps!!

6 0
3 years ago
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