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Gwar [14]
3 years ago
13

Aluminum oxide is formed when aluminum combines with oxygen in the air. How many grams of Al2O3 are formed when 13.6 mol of Al r

eacts?
4 Al + 3 O2 → 2 Al2O3
Chemistry
1 answer:
LenKa [72]3 years ago
3 0

Answer:

4Al + 3O2 → 2Al2O3 a. If you use 2.3 moles of Al, how many moles of Al2O3 can ... How many grams of Al2O3 are produced from the reaction of 5 moles of Al?

Explanation:

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What is created when an acid is mixed with a base?
Arturiano [62]

Answer:

If you mix equal amounts of a strong acid and a strong base, the two chemicals essentially cancel each other out and produce a salt and water. Mixing equal amounts of a strong acid with a strong base also produces a neutral pH (pH = 7) solution.

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3 years ago
What would the world be like without elements? please​
Lisa [10]

Answer:

It would be nothing. Quite literally nothing. No Oxygen, no dirt, no anything.

Explanation:

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2 years ago
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A bulldozer does 15,000 J of work lifting dirt 15 m up to get it out of a hole. How much force did the bulldozer use to accompli
nika2105 [10]

Answer:

<h3>The answer is 3000 N</h3>

Explanation:

The amount of force can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{15000}{5}  \\

We have the final answer as

<h3>3000 N</h3>

Hope this helps you

7 0
2 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
What is one example of closed economy?
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