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dimulka
2 months ago
10

Hydrogen has three isotopes with mass numbers of 1, 2, and 3 and has an average atomic mass of 1.00794 amu. This information ind

icates that
1. equal numbers of each isotope are present
2. more isotopes have an atomic mass of 2 or 3 than of 1
3. more isotopes have an atomic mass of 1 than of 2 or 3
4. isotopes have only an atomic mass of 1
Chemistry
2 answers:
alisha [2.9K]2 months ago
8 0

Answer: It shows that the majority of isotopes have a mass number of 1 over 2 or 3.

Explanation:

The average atomic mass is calculated by summing the masses of each isotope weighted by their relative abundances.

The formula applied is:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Given three hydrogen isotopes: H-1, H-2, and H-3.

The average atomic mass is 1.00794 amu.

This value being very close to the mass of H-1 indicates its natural abundance is much greater than those of H-2 and H-3.

The abundance of H-1 is approximately 99.98%.

The combined abundances of H-2 and H-3 are about 0.02%.

Therefore, this data implies that isotopes with mass 1 are more numerous than those with masses 2 or 3.

VMariaS [2.9K]2 months ago
7 0

The answer is actually 3, believe me.

Explanation:

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13.3 g of benzene (C6H6) is dissolved in 282 g of carbon tetrachloride. What is the molal concentration of benzene in this solut
Tems11 [2777]

Answer:

0.605 molal

Explanation:

Molality indicates the solute amount in a specific solvent mass.

Let’s find the amount of benzene solute.

Mass of benzene = 13.3g

Molar mass of C6H6 = 12*6 +1*6 =72+7=78g/mol

Amount of benzene = mass/molar mass

                           =13.3/78

                          =0.1705mol

Molality = amount of solute/mass of solvent in kg

Mass of solvent = 282g = 0.282kg

Molality = 0.1705/0.282

    =0.605 molal

6 0
28 days ago
The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO
lorasvet [2795]

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

87.54 mol O2 x \frac{15}{85} = 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2

4 0
28 days ago
Read 2 more answers
Acetone major species present when dissolved in water
Alekssandra [3086]

The compound is acetone ( CH₃-CO-CH₃)


Explanation:


1) Acetone is represented as CH₃-CO-CH₃.


2) This is a molecule formed by covalent bonds.


3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.


In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.


This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).



3 0
1 month ago
N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t
lorasvet [2795]

Answer: The mole fraction of nitrogen is calculated to be 0.4615.


Explanation: In the mixture of nitrogen (N_{2}) and hydrogen (H_{2}), the mole ratio established is 1: 1.5.


At this juncture, we identify that (H_{2}) serves as the limiting reagent.


When 0.4 moles of (NH_{3}) are produced, it will require 0.4 moles of (N_{2}) along with 3.4 moles of (H_{2}).


The total amount of remaining (N_{2}) ends up being 0.6, while the amount of (H_{2}) left in the mixture is 0.3 moles.


The mole fraction of (N_{2}) is computed as 0.6 divided by the sum of 0.6, 0.4, and 0.3, resulting in 0.4615.

8 0
2 months ago
Read 2 more answers
Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c
KiRa [2933]

Answer:

[Cl⁻] = 0.016M

Explanation:

To begin, we analyze the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This indicates a solubility equilibrium, resulting in the formation of lead(II) chloride precipitate. The salt can dissociate as follows:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

Given that this is an equilibrium scenario, the Kps serves as the constant (Solubility product):

Kps = s. (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵. 1/4)

s = 0.016 M

3 0
1 month ago
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