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stepan [7]
4 years ago
11

Use this equqation for the problem:4Al+3O2 --> 2Al2O3 How many moles of oxygen are needed to completely react with 14.8 mol o

f Al?​
Chemistry
1 answer:
Lesechka [4]4 years ago
7 0

Each mole of butane needs 6.5 moles of oxygen, so 13 moles of oxygen is required for 2 moles of butane in a complete combustion

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What volume of oxygen in liters at STP wpuld be required for the combustion of 17.0 grams PH3?
Lynna [10]

First, we need to state the chemical equation for the combustion of PH3

4PH3\text{ + 8O2 }\to\text{ P4O10 }+6H2O

And the mass of PH3 is 17.0 grams and we need to know the moles.

In the periodic table, the atomic mass of the P (phosphorus) is 31 and the atomic mass of the H (hydrogen) is 1.

So, you sum the mass of P to the mass of H multiplied by 3 and you obtain this:

\text{mole}cu\text{lar weight of PH3=31}\cdot1+1\cdot3=34\text{ g PH3/mol}

With this data, we can search the moles of PH3:

7 0
1 year ago
No. free poijnts ,,,,,,,
arsen [322]
Yes free pointssssss
6 0
3 years ago
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Which of these is a covalent compound?<br><br> A. LiCl<br> B. MgO<br> C. AlCl3<br> D. CO
Fiesta28 [93]

Answer:

C. AlCl3 is a covalent compound.

Explanation:

It is formed by mutual sharing of electrons between two atoms each contributing equal number of electrons to the electron pair.

3 0
3 years ago
Two beakers are placed in a small closed container at 25 °C. One contains 284 mL of a 0.296 M aqueous solution of C6H12O6; the s
Katen [24]

Answer:

Final volume = 0.103M x446ml/0.184m = 250ml

Explanation:

As time passes, the volume of solutions in the second beaker decreases and that in the first beaker increases. If we wait long enough, the final volumes and concentration in the beakers would be,

First beaker

Final concentration = 0.184M

Final volume = 0.296M x 284ml/0.184 = 457ml

Second beaker

Final concentration = 0.184M

Final volume = 0.103M x446ml/0.184m = 250ml

6 0
3 years ago
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What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same t
kenny6666 [7]

Answer: 20.7 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system=-P\Delta V  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = 4.51\times 4.184kJ=18.9kJ    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

\Delta E=+1.79+(18.9)=20.7kJ

Thus \Delta E for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ

4 0
3 years ago
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