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3 years ago

How many moles are in 100.0 g of gold?

Chemistry

1 answer:

Savatey [412]3 years ago

7 0

Answer:

196.96655

Explanation:

The answer is 196.96655. I assume you are converting between grams Gold and mole. The molecular formula for Gold is Au. The SI base unit for the amount of substance is the mole.

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for a given input force, a ramp increases which of the following? A)height B)output force C)output work D) efficiency

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Answer - C) output work

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4 years ago

What is the empirical formula of .50 moles of S and 1.5 moles of O

Ira Lisetskai [31]

Partial/total moles= 0.5/2 & 1.5/2 it results to a ratio S:O where S=1 and O=3 the final answer will be SO3.. I hope that helped

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4 years ago

What arrangement of electrons would result in a nonpolar molecule

alexdok [17]

For an arrangement of electrons to be nonpolar in a molecule they need to have equal electron charges.

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3 years ago

Chromium has an atomic mass of 51.9961 u and consists of four isotopes, 50 Cr , 52 Cr , 53 Cr , and 54 Cr . The 52 Cr isotope ha

Troyanec [42]

Answer : The atomic mass of $_{24}^{53}\textrm{Cr}$ isotope is 52.8367 amu

Explanation :

We know that:

Total percentage abundance of the isotope = 100 %

Percentage abundance of $_{24}^{50}\text{Cr}\text{ and }_{24}^{53}\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%$

We are given:

Ratio of $_{24}^{50}\textrm{Cr}\text{ and }_{24}^{53}\textrm{Cr}$ isotopes = 0.4579 : 1

Percentage abundance of $_{24}^{50}\textrm{Cr}$ isotope = $\frac{0.4579}{(0.4579+1)}\times 13.84\%=4.37\%$

Percentage abundance of $_{24}^{53}\textrm{Cr}$ isotope = $\frac{1}{(0.4579+1)}\times 13.84\%=9.49\%$

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows: $\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the mass of $_{24}^{53}\textrm{Cr}$ isotope be 'x'

For $_{24}^{50}\textrm{Cr}$ isotope:

Mass of $_{24}^{50}\textrm{Cr}$ isotope = 49.9460 amu

Percentage abundance of $_{24}^{50}\textrm{Cr}$ = 4.37 %

Fractional abundance of $_{24}^{50}\textrm{Cr}$ isotope = 0.0437

For $_{24}^{52}\textrm{Cr}$ isotope:

Mass of $_{24}^{52}\textrm{Cr}$ isotope = 51.9405 amu

Percentage abundance of $_{24}^{52}\textrm{Cr}$ isotope = 83.79 %

Fractional abundance of $_{24}^{52}\textrm{Cr}$ isotope = 0.8379

or $_{24}^{53}\textrm{Cr}$ isotope:

Mass of $_{24}^{53}\textrm{Cr}$ isotope = x amu

Percentage abundance of $_{24}^{53}\textrm{Cr}$ isotope = 9.49 %

Fractional abundance of $_{24}^{53}\textrm{Cr}$ isotope = 0.0949

For $_{24}^{54}\textrm{Cr}$ isotope:

Mass of $_{24}^{54}\textrm{Cr}$ isotope = 53.9389 amu

Percentage abundance of $_{24}^{54}\textrm{Cr}$ isotope = 2.37 %

Fractional abundance of $_{24}^{54}\textrm{Cr}$ isotope = 0.0237

Average atomic mass of chromium = 51.9961 amu

Putting values in equation 1, we get:

$51.9961=[(49.9460\times 0.0437)+(51.9405\times 0.8379)+(x\times 0.0949)+(53.9389\times 0.0237)]\\\\x=52.8367amu$

Hence, the atomic mass of $_{24}^{53}\textrm{Cr}$ isotope is 52.8367 amu

4 0

4 years ago

Which of the following represents the number of moles of a solute in one liter of a solution?

OverLord2011 [107]

Answer:

Q1: 2. Molarity.

Q2: 2.51 x 10⁻³.

Explanation:

Q1:

The molarity is defined as the no. of moles of solute in one liter of solution.

It is can be calculated from the relation: M = (no. of moles of solute)/(Volume in L of the solution).

Q2:

∵ pH = - log[H₃O⁺]

∴ 2.60 = - log[H₃O⁺]

log[H₃O⁺] = -2.6.

∴ [H₃O⁺] = 2.51 x 10⁻³.

7 0

3 years ago

Read 2 more answers