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3 years ago

What is the main exception in the atomic radius? Explain.

Chemistry

1 answer:

yarga [219]3 years ago

7 0

Answer:

The atomic radius of atoms generally decreases from left to right across a period. There are some small exceptions, such as the oxygen radius being slightly greater than the nitrogen radius. Within a period, protons are added to the nucleus as electrons are being added to the same principal energy level.

hope this helps!

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Are bath bombs covalent or ionic bonds?

lubasha [3.4K]

Bath Bombs are colvalent bonds because they have both metals and non-metals before and after it makes contact with water.

4 0

2 years ago

how many H+ ions are in a solution with a pH of 2 and how many H+ ions are in a solution with a pH of 6?

LiRa [457]

Answer:

i) pH = 2

pH = -log(H+)

:- (H+) = 10^(-2)

:- (H+) = 0.01 M

ii) pH = 6

pH = -log(H+)

:- (H+) = 10^(-6)

:- (H+) = 0.000001 M

Explanation:

By definition: pH = -log(H+).

Given your pH, solve for the H+ using the the following log rule:

if a = (+/-) log (b) then

b = 10^((+/-) a).

Also remember unit of concentration is molar (M)

3 0

3 years ago

What volume of Co2 (carbon (iv) oxide)

hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

7 0

3 years ago

What is the ph of a 0.027 M KOH solution?

Monica [59]

First you calculate the concentration of [OH⁻] in <span>solution :

POH = - log [ OH</span>⁻]

POH = - log [ 0.027 ]

POH = 1.56

PH + POH = 14

PH + 1.56 = 14

PH = 14 - 1.56

PH = 12.44

hope this helps!

8 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.