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3 years ago

Compare fission and fusion reactions and explain the difference between their energy input and output.

2 answers:

8 0

Both the two are atomic responses that deliver vitality, however, the applications are not the same. Parting is the part of a substantial, flimsy core into two lighter cores, and the combination is the procedure where two light cores join together discharging tremendous measures of vitality.

sertanlavr [38]3 years ago

7 0

Answer :

Nuclear fusion : It is defined as a process which involves the conversion of two small nuclei to form a heavy nuclei and some amount of energy is also releases.

Nuclear fission : It is defined as a process which involves the conversion of a heavier nuclei (which is an unstable nuclei) into two or more small nuclei (which is a stable nuclei) and a large amount of energy is also releases.

The difference between the nuclear fusion and nuclear fission is that, the energy is given in the nuclear fusion reaction but the energy is releases in the nuclear fission reaction.

In nuclear fusion, a small amount of energy is released but in nuclear fission, more amount of energy is released.

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Choose all that apply.

Mashutka [201]

A and D would be correct

5 0

3 years ago

The pKa of lactic acid is 3.9. A lactate buffer will be useful from pH values ________. The pKa of lactic acid is 3.9. A lactate

Vedmedyk [2.9K]

Answer:

Explanation:

The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.

This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

pH = pKa + log [A⁻]/[HA]

thus

0.1 ≤ [A⁻]/[HA] ≤ 10

Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.

Now we are equipped to answer our question:

pH range = 3.9 +/- 1 = 2.9 through 4.9

7 0

3 years ago

What is the pH of a 0.28 M solution of ascorbic acid

Nady [450]

Hey there!

Values Ka1 and Ka2 :

Ka1 => 8.0*10⁻⁵

Ka2 => 1.6*10⁻¹²

H2A + H2O -------> H3O⁺ + HA⁻

Ka2 is very less so I am not considering that dissociation.

Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]

lets concentration of H3O⁺ = X then above equation will be

8.0*10−5 = [x] [x] / [0.28 -x

8.0*10−5 = x² / [0.28 -x ]

x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵

solve the quardratic equation

X =0.004693 M

pH = -log[H⁺]

pH = - log [ 0.004693 ]

pH = 2.3285

Hope that helps!

3 0

3 years ago

) How many moles are there in 2.00 x 10^19 molecules of calcium phosphate?

Tamiku [17]

Answer:

$\huge 3.322 \times {10}^{ - 5} \: \: moles \\$

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{2.00 \times {10}^{19} }{6.02 \times {10}^{23} } \\ \\ = 3.322 \times {10}^{ - 5} \: \: moles$

Hope this helps you

4 0

3 years ago

* The cost of table salt and table sugar is Rs 15 per kg.

Vilka [71]

Cost per mole

Table salt : Rs 0.878

Table sugar : Rs 23.63

<h3>Further explanation</h3>

Given

Cost table salt (NaCl) = 15/kg

Cost table sugar(sucrose-C12H22O11) = 69/kg

Required

cost per mole

Solution

mol of 1 kg Table salt(NaCl ,MW= 58.5 g/mol) :

$\tt mol=\dfrac{1000~g}{58.5}=17.09~mol=Rs~15\rightarrow 1~mol=Rs~0.878$

mol of 1 kg Table sugar(C12H22O11 ,MW= 342 g/mol) :

$\tt mol=\dfrac{1000}{342}=2.92~mol=Rs~69\rightarrow 1~mol=Rs~23.63$

6 0

2 years ago