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4 years ago

6. Identify the number of neutrons in one atom of silicon with an isotopic mass number of 28 amu.

Chemistry

1 answer:

Sholpan [36]4 years ago

5 0

Explanation:

The number of neutrons in one atom of silicon with an isotopic mass number of 28amu is 14.

An atom is made up three fundamental sub-atomic particles which are:

Protons, neutron and electrons

Protons are the positively charged particles in an atom

Neutrons do not carry any charges

Electrons are the negatively charged particles.

In the nucleus of an atom, both protons and neutrons can be found. They are the massive particle in an atom.

The mass number of an atom = number of protons + number of neutrons

Number of neutrons = mass number - number of protons.

From the periodic table, we know that silicon has an atomic number of 14.

The atomic number is the number of protons in an atom:

Number of neutrons = 28 - 14 = 14amu

In the atom, we will have 14amu of neutrons in a silicon atom.

learn more:

Number of neutrons brainly.com/question/2757829

#learnwithBrainly

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Need help asap with this chemistry if someone could help me

Burka [1]

Answer:

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

Structure Number Two would likely be the most stable structure.

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

The N atom is the one that is "likely" to be attracted to an anion. See explanation.

Explanation:

When calculating the formal charge for an atom, the assumption is that electrons in a chemical bond are shared equally between the two bonding atoms. The formula for the formal charge of an atom can be written as:

$\text{Formal Charge} \\ = \text{Number of Valence Electrons in Element} \\ \phantom{=}-\text{Number of Chemical Bonds} \\\phantom{=} - \text{Number of nonbonding Lone Pair Electrons}$ .

For example, for the N atom in structure one of the first question,

N is in IUPAC group 15. There are 15 - 10 = 5 valence electrons on N.
This N atom is connected to only 1 chemical bond.
There are three pairs, or 6 electrons that aren't in a chemical bond.

The formal charge of this N atom will be $5 - 1 - 6 = -2$ .

Apply this rule to the other atoms. Note that a double bond counts as two bonds while a triple bond counts as three.

Structure One:

N: -2
C: 0
O: +1

Structure Two:

N: 0
C: 0
O: -1

Structure Three:

N: -1
C: 0
O: 0.

In general, the formal charge on all atoms in a molecule or an ion shall be as close to zero as possible. That rules out Structure number one.

Additionally, if there is a negative charge on one of the atoms, that atom shall preferably be the most electronegative one in the entire molecule. O is more electronegative than N. Structure two will likely be favored over structure three.

Similarly,

All five C atoms: 0
All six H atoms to C: 0
N atom: +1.

Assuming that electrons in a chemical bond are shared equally (which is likely not the case,) the nitrogen atom in this molecule will carry a positive charge. By that assumption, it would attract an anion.

Note that in reality this assumption seldom holds. In this ion, the N-H bond is highly polarized such that the partial positive charge is mostly located on the H atom bonded to the N atom. This example shows how the formal charge assumption might give misleading information. However, for the sake of this particular problem, the N atom is the one that is "likely" to be attracted to an anion.

5 0

3 years ago

Find the equation of directrix of the parabola is given by 64x equals to -ySquare

ANEK [815]

Answer:

Given equation of parabola is

and

=64x ......(i)

The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.

On differentiating both sides of equation (i), we get

2 y

d y

=64

⇒

d y

and

Also, slope of the given line is −

∴−

and

⇒and=−24

From equation (i), (−24)

=64x⇒x=9

∴ the required point is (9,−24)

Explanation:

This is the correct answer you want

please follow the

5 0

3 years ago

What is true of all body except sex cells

Evgesh-ka [11]

They can all by seperrated or replicated.

4 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

When ethane (

LuckyWell [14K]

Answer:

9 Moles

Explanation:

C2H6 has 6 Hydrogens and Water Has 2 Hydrogens

so it takes 1 mole ethane to produce 3 moles water

1 Mole Ethane ----> 3 Moles Water so 3 ----> 9 moles

7 0

4 years ago