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3 years ago

Which kind of energy transformation occurs during skydiving?

Chemistry

2 answers:

nikitadnepr [17]3 years ago

5 0

First it is gravitational energy, then it changes to Kinetic energy during the fall.

vredina [299]3 years ago

5 0

2nd one gravitational to kinetic

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How many moles are in 1.20 x 10^21 atoms of Phosphorus?

Ludmilka [50]

Answer:

You would get 19.969 moles

Explanation:

8 0

2 years ago

CAN ANYONE ANSWER THIS ONE CHEMISTRY QUESTION PLZZZ!!!

xz_007 [3.2K]

Answer:

2.8 oxygen

Explanation:

8 0

3 years ago

Read 2 more answers

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

(a) Calculate the total volume (in liters) of air an adult breathes in a day. (b) In a city with heavy traffic, the air contains

Furkat [3]

Answer:

a) V air/day = 8640 L air an adult breaths / day

b) 0.0181 L CO intake a person / day

Explanation:

a) one average person has 12 breaths for min:

P = 12 breath/min

in each breath it take an average of 500 mL on air.

1 breath ≅ 500 mL air

⇒ 12 breath / min * 500mL air / breath = 6000 mL air / min

the average air volume per day of a person is:

⇒ Vair/day = 6000 mL air / min * (60 min / h) * ( 24 h / day ) = 8640000 mLair / day * ( L / 1000 mL)

⇒ V air / day = 8640 L / day

b) 2.1 E-6 L CO / L air * 8640 L air / day = 0.0181 L CO / day

4 0

2 years ago

how many molecule of carbon dioxide are needed to react with excess iron oxide to produce 11.6 g of iron

lesya692 [45]

Answer:

0.16 moles of Carbon

Explanation:

The balanced reaction equation:

$2Fe_{2}O_{3}$ + $3C$ → $4Fe$ + $3CO_{2}$ ↑

The mole ratio of Carbon to Iron is 3 : 4 (since Fe2O3 is in excess)

i.e 3 moles of C produces 4 moles of Fe.

If 1 mole of Fe - 55.8g of Fe

? moles - 11.6g of Fe

= $\frac{11.6}{55.8}$ = 0.208 moles

But 3 moles of C - 4 moles of Fe

? moles of C - 0.208 moles of Fe

= $\frac{3 *0.208}{4}$ = 0.16 moles of carbon.

I hope this explanation was clear and useful.

5 0

3 years ago