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zlopas [31]
3 years ago
12

One of the hopes for solving the world's energy problem is

Chemistry
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

2.803013439419911 × 10⁻¹² J

Explanation:

Mass defect = mass of reactant - mass of product

(2.0140 + 3.01605) - (4.002603 + 1.008665)

5.03005 - 5.011268 = 0.018782 amu

mass in Kg = mass (amu) × 1.66053892173 × 10⁻²⁷ kg

mass in kg = 0.018782 × 1.66053892173 × 10⁻²⁷ = 3.1188242027932 × 10⁻²⁹kg

E = Δm c² where c is the speed of light = 2.9979 × 10⁸m/s

E = 3.1188242027932 × 10⁻²⁹kg × (2.9979 × 10⁸m/s)² = 2.803013439419911 × 10⁻¹² J

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Calculate the number of moles found in 3.045x1024 atoms of helium.<br><br> PLS HELP
Kisachek [45]

Explanation:

so for this u have to use this equation where

Moles = number of particle/6.02×10^23

= 3.045 × 10^24/6.02×10^23

= 5.0581

write it to 3 S.F so 5.06 moles

4 0
2 years ago
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8
Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is pK_a  =7.82

Explanation:

From the question we are told

    The volume of base is  V_B = 26.mL = 0.0260L

     The pH of solution is  pH =  7.82

      The concentration of the acid is C_A = 0.1M

From the pH we can see that the titration is between a strong base and a weak acid

 Let assume that the the volume of acid is  V_A = 18mL= 0.018L

Generally the concentration of base

                    C_B = \frac{C_AV_A}{C_B}

Substituting value  

                     C_B = \frac{0.1 * 0.01800}{0.0260}

                    C_B= 0.0692M

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

           HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}

Now before the reaction the number of mole of base is  

            No \ of \ moles[N_B]  =  C_B * V_B

Substituting value  

                    N_B = 0.01300 * 0.0692

                         = 0.0009 \ moles    

                                 

Now before the reaction the number of mole of acid is  

            No \ of \ moles  =  C_B * V_B

Substituting value  

                    N_A = 0.01800 *0.1

                         = 0.001800 \ moles

Now after the reaction the number of moles of  base is  zero  i.e has been used up

    this mathematically represented as

                         N_B ' = N_B - N_B = 0

    The  number of moles of acid is  

             N_A ' = N_A  - N_B

                   = 0.0009\ moles

The pH of this reaction can be mathematically represented as

                 pH  = pK_a + log \frac{[base]}{[acid]}

Substituting values

                  7.82 = pK_a +log \frac{0.0009}{0.0009}

                  pK_a  =7.82        

                     

             

                                 

       

           

                     

8 0
3 years ago
How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2
Irina18 [472]

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

All you have to do is to create a ratio between the molecule and the oxygen atom.

5 moles of Mg3(PO4)2 (4x2 moles O/1 mole Mg3(PO4)2) = 40 moles of oxygen

5 0
3 years ago
What is the molarity of a solution that contains 1.1 moles of lithium in 0.5 liters of solution?
cluponka [151]
0.5 litres contain 1.1 moles
therefore 1 litre will have= 1*1.1/0.5
             molarity=2.2M
8 0
3 years ago
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The stability of atomic nuclei is related to the
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Ratio of neutrons & protons .
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