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3 years ago

(Please help)

Chemistry

1 answer:

Orlov [11]3 years ago

7 0

Answer:

1. Isotope with mass number 39.

2. 39.02g/mol

Explanation:

1. The most common isotope is the isotope in the higher proportion, that is:

Isotope with mass number 39.

2. The average atomic mass is the sum of the masses times their abundance. For the element X:

Average atomic mass:

38*0.0967 + 39*0.7868 + 40*0.1134 + 41*0.0031

= 39.02g/mol

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An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is

Juli2301 [7.4K]

Answer:

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

Explanation:

The gas ideal law is

PV= nRT (equation 1)

Where:

P = pressure

R = gas constant

T = temperature

n= moles of substance

V = volume

Working with equation 1 we can get

$n =\frac{PV}{RT}$

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

$\frac{m}{mw} =\frac{PV}{RT}$ or

$m =\frac{P*V*mw}{R*T}$ (equation 2)

The cylindrical container has a constant pressure p

The volume is the volume of a cylinder this is

$V =(pi)*r^{2}*h$

Where:

r = radius

h = height

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is $V =(pi)*r^{2}*h$

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:

$T =\frac{T_{1} + T_{O}}{2}$

Replacing these values in the equation 2 we get:

$m =\frac{P*V*mw}{R*T}$ (equation 2)

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

8 0

4 years ago

The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. The balanced chemical reaction equatio

Bogdan [553]

Answer:

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

H° H₂O = -285.8 kJ/mol

Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

6 0

3 years ago

A chemistry student needs 10.0g of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the

iogann1982 [59]

Answer: The volume of ethanolamine the student should pour out is $9.80cm^3$

Explanation:

To calculate the volume of ethanolamine we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanolamine = $1.02g/cm^3$

Mass of ethanolamine = 10.0 g

Volume of ethanolamine = ?

Putting values in above equation, we get:

$1.02g/cm^3=\frac{10.0g}{\text{Volume of ethanolamine}}\\\\{\text{Volume of ethanolamine}=\frac{10.0g}{1.02g/cm^3}=9.80cm^3$

Thus the volume of ethanolamine the student should pour out is $9.80cm^3$

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4 years ago

How will fragmentation affect the mass spectrum of hydrogen chloride?

Dima020 [189]

It depends ..........................................

7 0

3 years ago

Please help with my science test

Paladinen [302]

Answer:

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3 years ago