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lara31 [8.8K]
3 years ago
5

How many moles are in 1.50 x 10^23 atoms of F? (2 decimal places)

Chemistry
1 answer:
Illusion [34]3 years ago
3 0

Explanation:

a) 0.4998 mol

b) 0.249 mol

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The element lithium (Li) has 3 protons and 3 electrons. The element fluorine (F) has 9 protons and 9 electrons. An atom of the e
Naddik [55]

When an atom of the element lithium (Li) transfers an electron to an atom of the element fluorine (F), then the bond results between the atoms is ionic bond.

<h3>what is chemical bond? </h3>

A chemical bond is defined as the bond which holds atoms together in molecules.

Bonds arise due to the electrostatic forces present between positively charged atomic nuclei and negatively charged electrons.

<h3>Types of chemical bond</h3>
  1. Ionic bond
  2. Covalent bond
  3. Coordinate bond

<h3>What is Ionic bond ? </h3>

Ionic bond is defined as the transfer of electron from one atom to another atom.

Since, electron transfer from lithium to fluroine. Thus lithium get positive charge and fluorine occupy negative charge.

Thus, the bond form between lithium atom and fluorine atom is ionic bond.

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8 0
1 year ago
1. The atomic number of an element is 23 and its mass number is 56. a. How many protons and electrons does an atom of this eleme
Arte-miy333 [17]

Answer:

There are 23 protons and 23 electrons. We know this because atomic number denotes the number of protons and also electrons. There are 56 - 23 = 33 neutrons because atomic mass denotes the total number of protons and neutrons.

7 0
4 years ago
We kept adding pbi2 into water until no more will dissolve. what is the concentration of lead (ii) iodide in the solution
blsea [12.9K]

PbI(ii) ionization in the solution of PBI(ii) into water is:

<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻

If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.

Therefore,

<span>Ksp=<span>[Pb</span></span>²⁺][I-]²

Plugging the values:

1.4×10⁻⁸ = x ⋅ (2x)²

1.4×10⁻⁸ = 4x³

x³ = {1.4×10⁻⁸}÷4

x³ = 0.35 x 10⁻⁸

or 

x³ = 3.5 x 10⁻⁹

x = 1.51 x 10⁻³

Hence,

Concentration of iodide ions in the solution:

2x = 3.02 x 10⁻³ 

6 0
3 years ago
100.0 mL of 3.8M NaCN, the minimum lethal concentration of sodium cyanide in blood serum
9966 [12]

The given question is incomplete. The complete question is:

Calculate the number of moles and the mass of the solute in each of the following solution: 100.0 mL of 3.8 × 10−5 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum

Answer: The number of moles and the mass of the solute are 0.38\times 10^{-5} and 18.62\times 10^{-5}g respectively

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

3.8\times 10^{-5}M=\frac{n\times 1000}{100.0}

n=0.38\times 10^{-5}

n = moles of NaCN = \frac{\text {given mass}}{\text {Molar mass}}

0.38\times 10^{-5}=\frac{x}{49g/mol}

x=18.62\times 10^{-5}g

Thus the number of moles and the mass of the solute are 0.38\times 10^{-5} and 18.62\times 10^{-5}g respectively

6 0
3 years ago
What mass of benzene is cooled from 83.8 °C to 77.1 °C when 167 J of energy is transferred out of the system? (The specific heat
dexar [7]

Answer:

14.32g

Explanation:

Initial temperature = 83.8°C

Final temperature = 77.1°C

Temperature change, ΔT = 83.8°C - 77.1°C = 6.7

Heat, H = 167J

Specific heat, c = 1.740J/g °C

m = ?

All these parameters are related with the equation below;

H = mcΔT

m = H / cΔT

m = 167 /  (1.740 * 6.7)

m = 167 / 11.658 = 14.32g

3 0
3 years ago
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