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IgorC [24]
3 years ago
10

A high jumper jumps 2.09m. If the jumper has a mass of 72kg, what is his gravitational potential energy at the highest point of

the jump?
Chemistry
1 answer:
Furkat [3]3 years ago
4 0
Si Ko alam hehehehe sorrry men
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C12H22011+1202-->12CO2+11H20
kogti [31]

Answer:

0.185moles

Explanation:

Given parameters:

Volume of O₂  = 49.8L

Unknown:

Number of moles of sucrose required  = ?

Solution:

We can assume that the reaction takes place at standard temperature and pressure.

From this, we can find the number of moles of oxygen that reacted and extrapolate to that of sucrose.

Chemical equation;

           C₁₂H₂₂0₁₁  +  120₂   →   12CO₂   +   11H₂0

Number moles  = \frac{volume of gas}{22.4}   at STP

Number of moles of oxygen gas = \frac{49.8}{22.4}   = 2.22moles

           12 moles of oxygen gas combines with  1 mole of sucrose

       2.22 moles of oxygen gas will combine with \frac{2.22}{12}   = 0.185moles

4 0
3 years ago
(d) carbon monoxide
Irina18 [472]

Answer:

D. carbon monoxide

Hope this helps :)

6 0
3 years ago
A new substance is formed when the?
quester [9]

Answer:

A chemical reaction happens when substances break apart or combine to form one or more new substances.

Explanation:

hope its right.

4 0
3 years ago
To evaluate R in an Ideal gas Law (PV=nRT), the Volume of a gas is measured in ____ , while the temperature of a gas is measured
Vika [28.1K]
L (litter)
K (kelvin)
3 0
2 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of NH_{3} is as follows:.

          2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)

And, initially only NH_{3} is present.

The given data is as follows.

  P_{NH_{3}} = 2.3 atm at equilibrium

   P_{H_{2}} = 3 \times P_{N_{2}} = 0.69 atm

Therefore,

          P_{N_{2}} = \frac{0.69 atm}{3}

                        = 0.23 aatm

So, P_{NH_{3}} = 2.3 - 2(0.23)

                       = 1.84 atm

Now, expression for K_{p} will be as follows.

         K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}

           K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}

                      = \frac{0.23 \times 0.33}{3.39}

                     = 0.0224

or,           K_{p} = 2.2 \times 10^{-2}

Thus, we can conclude that  the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is 2.2 \times 10^{-2}.

6 0
4 years ago
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