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allochka39001 [22]
3 years ago
12

Can someone explain these?​

Chemistry
1 answer:
elena-s [515]3 years ago
6 0

Answer:

what??

Explanation?/

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In determining nuclear binding energy, what is Einstein's equation used to do?
natita [175]
Basically this is used in calculating the nuclear binding energy by converting the mass defect (calculated first) to energy and if we recall, Einstein's equation E=mc2 is the perfection equation to use because E=mc2 in which E represents units of energy, m represents units of mass, and c 2 is the speed of light squared. 
3 0
4 years ago
How much energy is released when 0.40 mol C6H6(g) completely reacts with oxygen? 2C6H6(g) + 15O2(g) 12CO2(g) + 6H2O(g) kJ
Sladkaya [172]

Answer:

-1268 kJ

Explanation:

The molar enthalpy of combustion of C6H6 is -3170 kJ/mol

2C6H6(g) + 15O2(g) → 12CO2(g) + 6H2O(g)

Therefore;

when 1 mole of C6H6 undergoes combustion, an energy of -3170 kJ/mol is released.

Therefore; for 0.40 mole

= 0.4 × -3170 kJ

= -1268 kJ

7 0
3 years ago
Question 5 of 5<br> Carbon has six protons. Which model shows a neutral atom of carbon?
Studentka2010 [4]

Answer:

its d

Explanation:

8 0
3 years ago
The following data was collected when a reaction was performed experimentally in the laboratory.
Reptile [31]

The maximum amount of AlCl_3 produced during the experiment can be find out by knowing the Limiting Reagent.

Here, The maximum amount of AlCl_3 produced during the experiment is  400 grams

<h3>What is Limiting reagent ?</h3>

The limiting reagent (or limiting reactant or limiting agent) in a chemical reaction is a reactant that is totally consumed when the chemical reaction is stopped.

Now, According to the question,

Given ;

Number of moles of Al(NO_3)_3 - 4 moles

Number of moles of NaCl - 9 moles

To Find - Maximum amount of AlCl_3 produced during the reaction.

Solution ;

complete reaction ;

Al(NO_3)_3  + 3NaCl -->   3NaNO_3 + AlCl_3

To find the maximum amount of AlCl_3 produced during the reaction,

we need to find the limiting reagent.

Mole ratio Al(NO_3)_3  = 4/1 = 4

Mole ratio NaCl = 9/3 = 3

Thus, NaCl is the limiting reagent in the reaction.

Now, 3 moles of NaCl produces 1 mole of   Al(NO_3)_3

9 moles of NaCl will produce = 1/3 x 9 = 3 moles.

Weight of   Al(NO_3)_3 = 3 x 133.34 = 400 grams

Thus, 400 grams of   Al(NO_3)_3  is the maximum amount of   Al(NO_3)_3 produced during the experiment.

Learn more about Limiting Reagent Here ;

brainly.com/question/20070272

#SPJ1

8 0
2 years ago
What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur
Sunny_sXe [5.5K]

                        Al                                               S

    1)            <u>  35.94  </u>    =1.33111                     <u>  64.06 </u>     = 2.001875

                       27                                               32

    2)          =    <u> 1.33111 </u>                                 = <u>  2.001875  </u>

                     1.33111                                         1.33111

                 

    3)         =   ( 1 ) × 2                                     =    ( 1.5 ) ×2

    4)         =  2                                                =   3

                     Empirical Formula = Al2S3

1) Divide the percentage given in question by the Relative Atomic Mass (RAM)

of the given elements.

2) When you find the answers of the first part of question, divide these once again but this time, by the lowest number you found in part 1.

3) and 4) Write down the values. If you get a decimal which is in between 0.3-0.7 (including the 0.3 and 0.7), you cannot make it a whole number by rounding of. Therefore, multiply the decimal with a whole number until you get a whole number as your answer. In this question, when you multiply 1.5 by 2, the answer is 3 which is a whole number. Multiply the other whole number by the same number as that you multiplied for 1.5. And use these numbers in part 4 to make the empirical formula which is Aluminium Sulfide (Al2S3)

Hope this helps you :)))

Please give this a brainliest

6 0
3 years ago
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