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WINSTONCH [101]
3 years ago
10

What is the molar mass, in grams, of a mole of an element equal to?

Chemistry
2 answers:
guajiro [1.7K]3 years ago
5 0

Answer:

The molar mass of a mole of an element equal to the atomic weight of the element in grams.

Explanation:

The mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu).

The standard atomic weight is the average of the relative atomic masses of all the isotopes of a chemical element weighted by their respective abundance on Earth. It reflects the prevalence of each natural isotope of an element.

A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142×10²³ atoms.

Hence, the molar mass, in grams, of a mole of an element equal to the atomic weight of the element.

Paha777 [63]3 years ago
3 0
The molar mass, in grams, of a mole for an element is equal to the atomic weight if the element.
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double-displacement reaction

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We have the chemical reaction:

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3 years ago
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
Taya2010 [7]

Answer:

Here's what I get  

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

                   H₂ +    I₂    ⇌ 2HI

I/mol·L⁻¹:    0.30   0.15         x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

 

7 0
3 years ago
At 700 K, the reaction below has an Kp value of 54. An equilibrium mixture at this temperature was found to contain 0.933 atm of
kati45 [8]

Answer:

See explanation below

Explanation:

In this case, we have the equilibrium reaction which is:

H₂ + I₂ <------> 2HI       Kp = 54

Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:

Kp = PpHI² / PpH₂ * PpI₂

Solving for the partial pressure of iodine:

PpI₂ = PpHI² / PpH₂ * Kp

Replacing the given values, we have:

PpI₂ = (2.1)² / 0.933 * 54

PpI₂ = 4.41 / 50.382

PpI₂ = 0.088 atm

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