In electroplating, how should the object that will be plated be connected in the electrolytic cell?

Which description of salt is a physical property?

Nonamiya [84]

Answer:

White being the color and coming in small grains.

Explanation:

Physical properties are something you can clearly see about the object.

6 0

2 years ago

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Gasoline has a density of 0.749 g/ml. how many pounds does 19.2 gallons of gasoline weigh?

Svetradugi [14.3K]

A good first step is writing the amount in terms of ml.

19.2 gallons = 72.68 L = 72680 ml

that would mean it weighs 0.749*720680g = 54437.32ml = 54.437 L

hope that helps :)

6 0

2 years ago

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What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

Electrons are found in energy levels around the nucleus of an atom. Energy levels can also be called ""shells."" The energy leve

melamori03 [73]

Answer:

The first electronic shell or energy level, having n= 1, called the K shell, is the closest to the nucleus.

Explanation:

In an atom, the sub atomic particles called electrons, revolve around the nucleus of the given atom at different energy levels, called shells.

The energy levels or the electronic shells are denoted by the principal quantum number (n) and has the following values n = 1, 2, 3, 4, 5, 6, 7; from the innermost shell to the outermost.

The electron shells are also labelled by the alphabets, K, L, M, N, O, P, and Q, from the innermost shell to the outermost.

Thus the first electronic shell having n= 1, called K shell, is the lowest energy level and is present closest to the nucleus.

Whereas, the last electronic shell having n =7, called Q shell, is the highest energy level and is farthest from the nucleus.

5 0

3 years ago

A solution of KMnO4 has an absorbance of 0.539 when measured in the colorimeter. Determine the concentration of the KMnO4 given

loris [4]

Answer:

Concentration of unknown solution is 0.0416 M

Explanation:

As we know

Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration

From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.

From the graph, we can obtain-

Y = 5.73 X – 0.0065

Absorbance = 0.232

0.232 = 5.73 X – 0.0065

X = 0.0416

Concentration of unknown solution is 0.0416 M

6 0

2 years ago