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11 days ago

In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?

Chemistry

1 answer:

lorasvet [956]11 days ago

7 0

The total atom count in the sample is obtained by aggregating the number of atoms from each element in it.

The number of gold atoms, $Au$ , found in $KAu(CN)_2$ , equals $6.66\times 10^{20}$ atoms (according to given data).

Based on the compound formula $KAu(CN)_2$ , it's evident that potassium and gold quantities are equivalent, while carbon and nitrogen quantities are double that amount.

Thus, the number of atoms of each element is as follows:

For potassium, $K$ , in $KAu(CN)_2$ equals $6.66\times 10^{20}$ atoms.

For carbon, $C$ , in $KAu(CN)_2$ equals $2\times6.66\times 10^{20}$ atoms, which totals $13.32\times 10^{20}$ .

For nitrogen, $N$ , in $KAu(CN)_2$ equals $2\times6.66\times 10^{20}$ atoms, which totals $13.32\times 10^{20}$ .

The total atoms in $KAu(CN)_2$ are calculated as: Number of gold, $Au$ + Number of potassium, $K$ + Number of carbon, $C$ + Number of nitrogen, $N$ .

Hence, the overall atom count is $6.66\times 10^{20}+6.66\times 10^{20}+13.32\times 10^{20}+13.32\times 10^{20}$ .

The total number of atoms amounts to $39.96\times 10^{20}$ atoms.

Consequently, the total atom quantity in $KAu(CN)_2$ is $3.996\times 10^{21}$ atoms.

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3 days ago

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Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7

KiRa [971]

The Δ H value for butane (g) is -124.7 kJ/mol.

The Δ H value for CO2 (g) is -393.5 kJ/mol.

The Δ H value for H2O (g) is -241.8 kJ/mol.

The mass of butane is 8.30 grams.

Butane has a molar mass of 58 g/mol.

Considering the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

To determine the Δ H° of the reaction:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

By substituting values, we find that

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, we will calculate how many moles of butane are in 8.30 grams.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Therefore, the total energy released during the reaction is given by,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Thus, the total heat released in the reaction is 380.14 kJ.

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6 days ago

Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr

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Answer:

The integer value of x in the hydrate is 10.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Molar concentration of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

$0.0366 M=\frac{n}{5.00 L}$

$n=0.0366 M\times 5 mol=0.183 mol$

Weight of hydrated sodium carbonate = n = 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol

$n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}$

$0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}$

$106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}$

By solving for x, we arrive at:

x = 9.95, approximating to 10

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15 hours ago

Write the chemical formula for iridium(III) nitride?

lorasvet [956]

Answer:

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Explanation:

The molar mass is 330.2335, in case that's also required.

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9 days ago

In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an ar

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Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in $15.3 \frac{atoms}{min-g}$

The dissolution rate of the sample is given by $9.16 \frac{atoms} {min-gram}$

The C-14 proportion within the sample can be determined as per $= \frac {9.16}{15.3} \\\\ = 0.5987$

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

$(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\$

$\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740\\$

Thus, the age of the sample is represented as = $n \times\ half-life$

$= 0.740 \times 5730 \ years \\\\=4241.17 \ years\\\\$

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1 day ago