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xz_007 [3.2K]
2 years ago
5

john has an object suspended in the air. It has a mass of 50 kilograms and is 50 meters above the ground. Calculate the objects

potential energy
Chemistry
1 answer:
egoroff_w [7]2 years ago
4 0
Answer is: <span>the objects potential energy is 24500 J.
</span>m(object) = 50 kg.
h(object) = 50 m.
g = 9,8 m/s².
E(object) = m·g·h.
E(object) = 50 kg · 9,8 m/s² · 50 m.
E(object) = 24500 N·m = 24500 J = 24,5 kJ.
g - <span>the acceleration of free fall.
mg - </span><span>weight of the object.</span>
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A 10 gram sample of iron reacts with oxygen to form 18.2 grams of ferric oxide. How many grams of oxygen reacted?
RoseWind [281]

Answer:

\boxed {\boxed {\sf 8.2 \ grams}}

Explanation:

According to the Law of Conservation of Mass, the mass of the products must equal the mass of the reactants.

  • mass products = mass reactants

In this problem, the reaction is:

iron + oxygen = ferric \ oxygen

  • The reactants are iron and oxygen. We know the mass of the iron sample is 10 grams.
  • The product is ferric oxide. The mass of the ferric oxide sample is 18.2 grams.

10 \ g + oxygen=18.2 \ g

We want to find how many grams of oxygen reacted. We have to get the oxygen by itself. 10 is being added to oxygen. The inverse of addition is subtraction. Subtract 10 from both sides of the equation.

10 \ g - 10 \ g+ oxygen = 18.2 \ g - 10 \ g

oxygen= 18.2 \ g - 10 \ g

oxygen= 8.2 \ g \\

<u>8.2 grams of oxygen </u>reacted with 10 grams of iron to form 18.2 grams of ferric oxide.

8 0
2 years ago
When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g
Serga [27]

<u>Answer:</u> The mass of iron produced will be 77.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For FeO:</u>

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol

  • <u>For aluminium:</u>

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol

The given chemical reaction follows:

3FeO+2Al\rightarrow 3Fe+Al_2O_3

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = \frac{3}{2}\times 0.93=1.395mol of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = \frac{3}{2}\times 0.93=1.395moles of iron metal

  • Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g

Hence, the mass of iron produced will be 77.6 grams

4 0
3 years ago
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 test tube contains a pure colourless gas. It is either oxygen,carbon dioxide,hydrogen or nitrogen. When a burning splint is pu
tekilochka [14]

The next test would be the lime water test.

Explanation:

If placing the gas in lime water will turn the solution milky, then the gas is most probably carbon dioxide, the lime water (Ca(OH)₂) turns milky due to the formation of solid CaCO₃ that is insoluble after reaction with carbon dioxide.

Carbon dioxide also puts out the flame in the splint test because it does not support combustion.

4 0
3 years ago
What is the density of a 32g toy car that<br> has a volume of 8cm3?<br> A. 256alcm3<br> An3
Greeley [361]

Answer:

density=mass/volume

density=32/8

density=4

8 0
3 years ago
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