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4 years ago

7

1. Given the hypotenuse of a 45°, 45°, 90° triangle is 8, find the length of the legs. Leave your answers in radical form if nee

ded.
453
legs =

1 answer:

Alex17521 [72]4 years ago

5 0

Answer:

5.66

Explanation:

sin = opposite/hypotenuse

8 sin = opposite

8 sin 45 = 5.66

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2 years ago

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By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an

Serggg [28]

Answer:

The Kinetic Energy is approximately 3 times decreased

Explanation:

A baseball weighs 5.13 oz.

a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?

b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.

Kinetic Energy (KE)=0.5×mass×velocity ^ 2

Kinetic Energy (KE)=0.5×mass × velocity ^ 2

Joules = kg×m^2/s^2

1 mile = 1609.344 meters

1 hour = 3600 sec

1 Oz = 28.34952 g = 0.02834952 kg

a) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=130.761 kg×m^2/s^2 = 130.761 Joules

b) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=43.51028 kg×m^2/s^2 = 43.51028 Joules

= 130.761 / 43.51028 = 3.00528,

As such the Kinetic Energy is approximately 3 times decreased

4 0

4 years ago

Identify and describe 2 changes of state that release energy

soldier1979 [14.2K]

Answer: All changes of state involve the transfer or energy

Explanation: i got my information from this site

https://www.esrl.noaa.gov/gmd/education/info_activities/pdfs/CTA_the_water_cycle.pdf

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3 years ago

Write the equilibrium constant expression for the reaction below.

kodGreya [7K]

Answer:

Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]

Explanation:

The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.

It works with molar concentrations on the equilibrium and it does not consider the solids compounds

Kc also can be modified by the time of the reaction.

This reaction is:

CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)

Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]

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3 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = $\frac{20}{125}$ = 0.16M

<u>Molarity of H20</u>:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

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3 years ago