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3 years ago

Which type of bond is present in hydrogen sulfide (H2S)? The table of electronegativities is given

Chemistry

2 answers:

LuckyWell [14K]3 years ago

7 0

Bonds formed between atoms can be classified as ionic and covalent

Ionic bonds are formed between atoms that have a high difference in the electronegativity values.

In contrast, bonds formed between atoms that have a difference in electronegativity lower than the ionic counterparts are polar covalent bonds. If the atoms have very similar electronegativities, they form non-polar covalent bonds.

In H2S, the S atom is bonded to 2 H atoms. The electronegativity of H = 2.2 and S= 2.56. Since the difference is not high the bond formed will be covalent (polar covalent).

Alexxandr [17]3 years ago

4 0

Answer:

nonpolar covalent

Explanation:

is answer on plato

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Wet snow is best for snow man building<br>

fredd [130]

Answer:

<h2>Hope my answer helps</h2>

Explanation:

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2 years ago

Calculate the solubility of carbon dioxide at 400 kPa.

BaLLatris [955]

The solubility of carbon dioxide at 400 kPa at room temperature is ;

( B ) 0.61 CO2/L

<u>Given data </u>

pressure of CO₂ = 400 Kpa = 3.95 atm

Kh of CO₂ = 3.3 * 10⁻² mol/L.atm

<h3>Calculate the solubility of carbon dioxide </h3>

Solubility = pressure * Kh value of CO₂

= 3.95 atm * 3.3 * 10⁻² mol / L.atm

= 0.13 mol/l CO₂

= 0.61 CO₂ / L

Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l CO₂.

Learn more about solubility : brainly.com/question/23946616

From the options the closest answer is ( B ) 0.61 CO₂ / L

7 0

2 years ago

During the apollo missions,an astronaut took a golf ball to the moon. He placed it on the moon’s surface and hit the ball with a

aleksandr82 [10.1K]

Answer:

b.)

Explanation:

Because you would have a different weight on the moon because of it's low gravity

3 0

3 years ago

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

UNO [17]

Answer: 0.100 m $K_2SO_4$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0$ = Elevation in boiling point

i= vant hoff factor

$K_b$ = boiling point constant

m= molality

1. For 0.100 m $K_2SO_{4}$

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be $3\times 0.100=0.300$

2. For 0.100 m $LiNO_3$

$LiNO_3\rightarrow Li^{+}+NO_3^{-}$

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be $2\times 0.100=0.200$

3. For 0.200 m $C_2H_8O_3$

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be $1\times 0.200=0.200$

4. For 0.060 m $Na_3PO_4$

$Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}$

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be $4\times 0.060=0.24m$

Thus as concentration of solute is highest for $K_2SO_4$ , the elevation in boiling point is highest and thus has the highest boiling point.

3 0

3 years ago

How many grams of hydrochloric acid are produced when 15.0 grams NaCl react with excess H2SO4 in the reaction

e-lub [12.9K]

Answer:

11.65 g

Explanation:

Amount of NaCl = 15.0 grams

amount of H₂SO₄ = Excess

mass of hydrochloric acid (HCl) = ?

Solution:

To solve this problem first we will look for the reaction that NaCl react with H₂SO₄

Reaction:

2NaCl + H₂SO₄ --------> 2 HCl + Na₂SO₄

As the H₂SO₄ is in excess so the amount of hydrochloric acid (HCl) depends on the amount of NaCl as it its act as limiting reactant.

Now if we look at the reaction

2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

1 mol 2 mol

Now convert moles to mass

Then

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

So,

2NaCl + H₂SO₄ --------> 2HCl + Na₂SO₄

2 mol (58.5 g/mol) 2 mol (36.5 g/mol)

94 g 73 g

So if we look at the reaction; 94 g of NaCl gives with 73 g of hydrochloric acid (HCl) in a this reaction, then how many grams of hydrochloric acid (HCl) will produce from 15.0 g of NaCl

For this apply unity formula

94 g of NaCl ≅ 73 g of HCl

15.0 g of NaCl ≅ X g of HCl

By Doing cross multiplication

X g of HCl = 73 g x 15.0 g / 94 g

X g of HCl = 11.65 g

11.65 g of hydrochloric acid (HCl) will produce by 15.0 g of NaCl

7 0

3 years ago