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Svet_ta [14]
2 years ago
13

Which of the following statements best describes the difference between

Chemistry
1 answer:
EastWind [94]2 years ago
6 0

Answer:

The correct answer is option A. "More energy is produced per reaction in a fusion reaction than in a fuel cell".

Explanation:

It is estimated that a fusion reaction four million times more energetic than traditional chemical reactions such as burning of coal, oil or gas. Fusion reactions produce even more energy than other types of atomic approaches, such as the ones used in hydrogen fuel cells. Hydrogen fuel cells are based on the energy produced by hydrogen atoms moving from an anode to a cathode, an energetic approach but not as energetic as the atomic fusion.

You might be interested in
In a reaction, 25 grams of reactant
Vsevolod [243]

Answer:

D. 15g

Explanation:

The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.

In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.

Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.

8 0
2 years ago
Consider NH3 and PH3. Electronegativities: P = 2.1, H = 2.1, N=3.0. Which statement is false?
NikAS [45]

Answer:

The P-H bonds are more polar than the N-H bonds.

Explanation:

Phosphine is a polar molecule with non-polar bonds. The phosphorus atom is bonded to three hydrogen atoms and the phosphorus atom has a lone pair of electrons. Since hydrogen and phosphorus are equal in electronegativity, it implies that they attract the shared pairs of electrons the same amount,hence bonding electrons are shared equally making the covalent bonds non-polar.

The lone pair of electrons on phosphorus causes the molecule to be asymmetrical with respect to charge distribution this is why the molecule is polar even though the are non-polar bonds in the molecule.

Looking at the values of electro negativity stated in the question, one can easily see that the difference in electro negativity between nitrogen and hydrogen is 0.9 while the difference in electro negativity between phosphorus and hydrogen is zero. It is clear that NH3 is naturally more polar than PH3 since each individual N-H bond in NH3 is a polar bond while the individual P-H bonds in PH3 are nonpolar.

4 0
2 years ago
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the
DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × 10^{-46} KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × 10^{-18}(\frac{1}{n^{2}_{final} } - \frac{1}{n^{2}_{initial}  })

    (1/434 × 10^{-9}) = -2.178 × 10^{-18} (\frac{n^{2}_{initial} - n^{2}_{final}  }{n^{2}_{final} n^{2}_{initial}   })

⇒ 434 × 10^{-9} = (1/-2.178 × 10^{-18})\frac{n^{2}_{final} *n^{2}_{initial}   }{n^{2}_{initial} - n^{2}_{final}    }

But, n_{final} = 2

434 × 10^{-9} = (1/2.178 × 10^{-18})\frac{2^{2} n^{2}_{initial}  }{n^{2}_{initial} - 2^{2}  }

434 × 10^{-9}  × 2.178 × 10^{-18} = (\frac{4n^{2}_{initial}  }{n^{2}_{initial} - 4 })

⇒ n_{initial} = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × 10^{8} ) ÷ (434 × 10^{-9})

    = (1.9878 × 10^{-25}) ÷ (434 × 10^{-9})

    = 4.58 × 10^{-19} J

    = 4.58 × 10^{-22} KJ

But 1 mole = 6.02×10^{23}, then;

energy in KJ/mole = (4.58 × 10^{-22} KJ) ÷ (6.02×10^{23})

         = 7.61 × 10^{-46} KJ/mole

7 0
2 years ago
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K
Dmitry [639]

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

5 0
2 years ago
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