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gladu [14]
3 years ago
14

How many atoms are in one mole of gold? A) 1.97 x 1023 atoms. B) 6.02 x 1023 atoms C) 79 atoms. D) 197 atoms

Chemistry
2 answers:
Hatshy [7]3 years ago
8 0

Answer: B) 6.022\times10^{23} atoms.

Explanation:

According to the International System of units ,

A mole is basically denotes a unit that contains 6.022\times10^{23} of atoms.

It is also known as <em>Avagadro Number.</em>

Similarly,  the number of atoms in one mole of gold = 6.022\times10^{23}

hence, the correct answer is B) 6.022\times10^{23} atoms.

Kobotan [32]3 years ago
3 0

B. Is the Answer 6.022 * 1023


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Answer:

Ionic bonds form when atoms transfer electrons between each other, forming ions that are electrically attracted to each other forming a bond between them. Sodium chloride (NaCl) is a typical ionic compound. The picture below shows both a sodium and a chlorine ion.

Explanation:

ionic bond

Because of the propensity of sodium to lose an electron and of chlorine to gain an electron, the elements are well suited to bond with one another. This transfer of electrons results in the formation of the ionic bond holding Na+ and Cl– together.

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How do CFC’s illustrate that technical advances can be a “double-edged” sword?
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Answer:

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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