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alexandr1967 [171]
3 years ago
9

How many neutrons are in a typical silicon atom? A. 15 B. 14 C. 28 D. 42

Chemistry
1 answer:
Aleksandr [31]3 years ago
7 0
Silicon (Si) atomic mass is 28.085.
To find the mass number of most common isotope we need to round atomic mas to wholes.
28.085≈28 is atomic mass
Atomic number(number of protons) is an ordinal number of an element in the Periodic Table.
 For silicons it is 14.
Mass number=number of protons(atomic number)+number of neutrons.
Number of neutrons=Mass number - Number of Protons(atomic number)
Number of neutrons= 28 - 14 = 14

Correct answer is B.14

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mamaluj [8]

Atoms are indivisible spheres-Dalton model

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Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0
3 years ago
During a titration a student found that 20.0cm3 of sodium carbonate solution
vlada-n [284]

Answer: The concentration of Na_2CO_3 required is 0.349mol/dm^3

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity of HNO_3 = 1

M_1 = molarity of HNO_3 solution = 0.500 mol/dm^3

V_1 = volume of  HNO_3 solution = 27.9cm^3

n_2 = acidity of Na_2CO_3 = 2

M_1 = molarity of Na_2CO_3 solution =?

V_1 = volume of  Na_2CO_3 solution =  20.0cm^3  

Putting in the values we get:

1\times 0.500\times 27.9=2\times M_2\times 20.0

M_2=0.349mol/dm^3

Therefore, concentration of Na_2CO_3 required is 0.349mol/dm^3

4 0
3 years ago
what mass of KCI ( solubility= 34.0 KCI/100 g water at 20 degrees celsius) can dissolve in 3.0x 10² g of water?
Alja [10]

Answer:

6g/mol

Explanation:

5 0
3 years ago
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