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3 years ago

Which substance would be most suitable for row 3?

1 answer:

7 0

The substance that would be most suitable for row 3 is SODIUM CHLORIDE.
The periodic table is made up of groups and periods, the periods are also called rows. There are seven rows and 18 groups in the periodic table. The third period is made up of 8 elements; both sodium and chlorine are part of the eight elements, thus, sodium chloride is more likely to be found on the third row.

You might be interested in

Why is the formula for CO2 Carbon Dioxide<br> and not Monocarbon Dioxide? Explain.

bekas [8.4K]

Answer:

If there is only one atom for the first element in the chemical formula, the prefix mono is typically omitted from the name. For example, CO is carbon monoxide, not mono carbon monoxide. ... Thus the chemical formula for carbon dioxide is CO2, not C1O2.

Explanation:

3 0

3 years ago

8KCIO3 +1C12H22011

docker41 [41]

3.2 g KClO3

Explanation:

1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)

= 0.0032 mol C12H22O11

0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)

= 0.026 mol KClO3

Therefore, the minimum amount of KClO3 needed is

0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)

= 3.2 g KClO3

5 0

3 years ago

elements are arranged in groups by similar atomic structure on the periodic table. this allows for an elements properties to be

emmasim [6.3K]

Answer:

SUB TO HAVIN RX

Explanation:

8 0

3 years ago

A common car battery consists of six identical cells each of which carries out the reaction: Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 +

docker41 [41]

Answer:

The correct answer is 4.58 grams.

Explanation:

Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).

For the reaction, n * 96500 C = molar mass

1C = molar mass/n*96500 = Equivalent wt / 96500

w = Equivalent wt / 96500 * I * t

In the given reaction,

Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.

Now putting the values in the equation we get,

w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)

w = 4.58 gm.

6 0

4 years ago

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago