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Anna35 [415]
3 years ago
13

Element Q has an atomic number 68. Consider the isotope Q-136 how many protons are in a neutral atom of this isotope?

Chemistry
1 answer:
mote1985 [20]3 years ago
7 0

Answer: 68

Explanation:

Isotopes of an element have same number of protons but different number of neutrons. Which means isotopes of an element have same atomic number but different mass number.

Atomic number is equal to the number of protons or the number of electrons for a neutral atom and is specific to a particular element.

Mass number is the sum of number of protons and the number of neutrons.

Given : atomic number of element Q = 68 = number of protons

Mass number of isotope Q-136 = 136

But as isotopes have same atomic number, the number of protons will be same and hence there are 68 protons are in a neutral atom of this isotope.

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Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
If the edge length of the unit cell is 705.2 pm, what is the density of KI in g/cm3.
Kipish [7]

The density  is 3.144 g / cm^3.

<u>Explanation</u>:

If effective number of atom in NaCl type structure, z = 4

a = 705.2 pm ⇒ In centimeter = 705.2 \times 10^-10

Na = 6.023 \times 10^23

density = (molecular weight) (z) / (Na) (a^3)

where molecular weight of KI is 166 g,

           Z represents the atomic number

density = (molecular weight) (z) / (Na) (a^3)              

             = (166 \times 4) / (6.023 \times 10^23) \times (705.2 \times 10^-10)

density = 3.144 g / cm^3.

5 0
3 years ago
Comparison between alkyl nucleophilic substitution with acyl nucleophilic substitution
S_A_V [24]

Answer:

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4 0
3 years ago
g A mass spectrum. The peak at mass 100 has a 8% relative abundance. THe peak with mass 85 has a 40% abundance. The peak at 71 h
Sati [7]

Answer:

The peak at mass 100 with a 8% relative abundance is the molecular ion peak

Explanation:

Molecular ion peak has the highest charge to mass ratio,

Mass of 100 is same as mass to charge ratio =100

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3 years ago
What word or two-word phrase best describes the shape of the nitrite ( NO2- ) ion?
Olin [163]

I think the answer would be nitrogen

6 0
3 years ago
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