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3 years ago

How many total protons neutrons and electrons are in an atom of iodine with a mass number of 127

1 answer:

8 0

The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is 74 (127 − 53 = 74).... i think....

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago

Where is the mass of an atom found

Virty [35]

Answer:

The mass of an atom is found in its nucleus.

Explanation:

An atom is made of three different particles: protons, neutrons and electrons.

Protons (positive charge) and neutrons (no charge) each have a mass of 1 AMU. They are both found in the nucleus (centre) of the atom.

Electrons (negative charge) are considered to have a mass of 0. Their mass is not actually 0, but very close so we do not count them. They are not in the nucleus, but found in shells surrounding the atom.

To calculate the mass of an atom, we add the number of protons and the number of electrons.

m = P + N

5 0

3 years ago

Which equation represents a double replacement reaction?

AveGali [126]

Answer:

C) LiOH + HCl → LiCl + H₂O

General Formulas and Concepts:

Chemistry - Reactions

Synthesis Reactions: A + B → AB
Decomposition Reactions: AB → A + B
Single-Replacement Reactions: A + BC → AB + C
Double-Replacement Reactions: AB + CD → AD + BC

Explanation:

Step 1: Define

RxN A: 2Na + 2H₂O → 2NaOH + H₂

RxN B: CaCO₃ → CaO + CO₂

RxN C: LiOH + HCl → LiCl + H₂O

RxN D: CH₄ + 2O₂ → CO₂ + 2H₂O

Step 2: Identify

RxN A: Single Replacement Reaction

RxN B: Decomposition Reaction

RxN C: Double Replacement Reaction

RxN D: Combustion Reaction

7 0

3 years ago

35.0 mL of acid with an unknown concentration is titrated with 24.6 mL of 0.432 M base. What is the concentration of the acid? A

dusya [7]

Given:

35.0 mL of acid with an unknown concentration

24.6 mL of 0.432 M base

Required:

Concentration of the acid

Solution:

M1V1 = M2V2

M1 (35.0 mL of acid) = (0.432 M base) (24.6 mL of base)

V1 = (0.432 M base) (24.6 mL of base) / (35.0 mL of acid)

M1 = 0.304 M of acid

5 0

3 years ago

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade

egoroff_w [7]

Answer:

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

Explanation:

Assuming Pressure is constant.

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

$\frac{300}{273} = \frac{V_2}{288}$

$V_2 = \frac{288*300}{273} \\V_2 = 316.5$

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

5 0

3 years ago