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Alexxx [7]
3 years ago
12

The diagram below is an artist’s impression of a single atom of element Be. The neutrons are shown with stripes, the protons are

gray, and the electrons are black. These particles are not drawn to scale. mc017-1.jpg Which element below could be an isotope of this atom? sodium-10 beryllium-10 boron-9 carbon-9

Chemistry
2 answers:
Ahat [919]3 years ago
3 0

Answer is: beryllium-10.

The diagram shows that atom has 4 protons, 5 neutrons and 2 valence electrons.

Atomic number is the number of protons, which is characteristic of a chemical element, beryllium (Be) is an element with atomic number 4.

Two valence electrons means that atom is from 2. group of periodic table, only beryllium is from that group; sodium (1. group), boron (13. group) and carbon (14. group).

Beryllium-10 has 6 neutrons, so it is isotope (different number of neutrons or mass number).

hjlf3 years ago
3 0

Answer: B

Explanation:

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Explanation:

To find the moles of 17.2 g K₂S, we need to know the molar mass to convert.

17.2g*\frac{mol}{110.256 g} =0.156 mol

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6 0
3 years ago
Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients
Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • First balance the main element in the reaction.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • Now balance oxygen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-+3H_2O

  • Now balance hydrogen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-+6H^+\rightarrow Br^-+3H_2O

  • Now balance the charge.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}+2e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : 3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The balanced chemical equation in acidic medium will be,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

7 0
3 years ago
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
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