All categories

1 month 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 [2.7K]1 month 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.

You might be interested in

Cathodic protection of a metal pipe against corrosion usually entails __________.

VMariaS [2998]

Answer:

C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.

Explanation:

Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.

This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.

3 0

1 month ago

The ionic radius of a sodium ion is 2.27 angstroms (A) . What is this length in um

lions [2927]

$\boxed{\sf 1Å=10^{-10}m}$

$\\ \rm\longmapsto 2.27Å$

$\\ \rm\longmapsto 2.27\times 10^{-10}m$

$\\ \rm\longmapsto 0.227\times 10^{-9}m$

$\\ \rm\longmapsto 0.0227\times 10^{-8}m$

$\\ \rm\longmapsto 0.00227\times 10^{-7}m$

$\\ \rm\longmapsto 0.00023\times 10^{-6}m$

$\\ \rm\longmapsto 0.00023\mu m$

6 0

1 month ago

Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Hydroch

VMariaS [2998]

Answer:

The categorization of strong, weak, and non-electrolytes is detailed below, based on the examples presented in the question.

Explanation:

A strong electrolyte fully dissociates or nearly so in an aqueous environment; typically, strong acids, bases, and salts fall under this category. Examples of strong electrolytes include:

Hydrochloric acid, HCl
Calcium hydroxide, Ca(OH)2
Potassium chloride, KCl

A weak electrolytepartially ionizes in solution; weak acids and bases are primary instances. Examples consist of:

Methylamine, CH3NH2
Hydrofluoric acid, HF

A non-electrolytedoes not dissociate in an aqueous medium. Examples of non-electrolytes are:

Sucrose, C12H22O11
Methanol, CH3OH

5 0

1 month ago

Read 2 more answers

n the table below, write the density of each object. Then predict whether the object will float or sink in each of the fluids. W

Anarel [2989]

Answer:

0.5 g/mL----- will float

1.0 g/mL---- will float

2.0 g/mL----- will sink

Explanation:

Objects with a density less than or equal to that of water will float due to having a lower mass, while objects with a density exceeding that of water will sink because their mass is greater than that of water. Thus, objects with a density of 0.5 g/mL and 1.0 g/mL will float since they are less dense than water (1 g/mL), whereas an object with a density of 2.0 g/mL will sink.

3 0

1 month ago

32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?

alisha [2963]

At standard temperature and pressure, it is established that 1 mole of gas has a volume of 22.4 liters.

According to the periodic table:
the molar mass of oxygen is 16 g
and the molar mass of hydrogen is 1 g
Hence, the molar mass of water vapor is calculated as 2(1) + 16 = 18 g

Thus, 18 g of water occupies 22.4 liters, therefore:
the volume for 32.7 g is (32.7 x 22.4) / 18 = 40.6933 liters

5 0

1 month ago