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pishuonlain [190]
3 years ago
15

Write the balanced equation showing hydrogen iodide (HI(g)) dissolving in water.

Chemistry
1 answer:
Andreas93 [3]3 years ago
6 0
According to Bronsted-Lowery concept of acid and base, Acid is that substance which donated H⁺, while, Base is that substance which accepts H⁺.

Hydroiodic Acid being a strong Acid with Ka value of 3.2 × 10⁹ is when dissolved in water it donates H⁺ to water and water acts as a Base. i.e. 

                                HI  +  H₂O    →    H₃O⁺  +  I⁻

H₃O⁺ formed is the conjugate Acid of H₂O, while I⁻ is the Conjugate Base of HI.
You might be interested in
Is more chemical weathering likely to occur in New York location or in the Illinois location? Explain your answer. HELP NEEDED (
Naily [24]
The pH of pure water is 7.0, which is neutral.

For a pH below this, the water is acidic. Substances that are acidic are often corrosive and thus could cause weather damaging. The pH of the precipitation in NY is below that of the precipitation in IL, and NY receives more precipitation, so for both of those reasons, it is likely to have more chemical weathering.
3 0
3 years ago
How many grams are in 5.2 moles of Li2SO4
skelet666 [1.2K]

Answer:

572 g

Explanation:

Molar mass is the mass of 1 mol of an element or compound

molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound

molar masses of each element making up lithium sulphate

Li - 7 g/mol

S - 32 g/mol

O - 16 g/mol

molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )

molar mass = 110 g/mol

mass of 1 mol of Li₂SO₄ is 110 g

therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g

mass is 572 g

7 0
3 years ago
What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
3 years ago
Calculate the amount of copper (in moles) in a 45.2 g pure copper sheet
tangare [24]

Answer:

0.712 mol

Explanation:

The easiest way to do this is to use a proportion.

1 mol of copper = 63.5 grams (check this using your periodic table).

x mol of copper = 45.2 grams

1/x = 63.5 / 45.2              Cross multiply

63.5 x = 1 * 45.2              Divide by 63.5

x = 45.2/63.5

x = 0.712 mol                   Answer to 3 sig digs

8 0
3 years ago
A 0.200 g sample of unknown metal x is dropped into hydrochloride acid and realeases 80.3 mL of hydrogen gas at STP using ideal
s344n2d4d5 [400]

Answer:

The number of mole of the unknown metal is 3.58×10¯³ mole

Explanation:

We'll begin by calculating the number of mole hydrogen gas, H2 that will occupy 80.3 mL at stp.

This is illustrated below:

Recall:

1 mole of any occupy 22.4L or 22400 mL at stp.

1 mole of H2 occupies 22400 mL at stp.

Therefore, Xmol of H2 will occupy 80.3 mL at stp i.e

Xmol of H2 = 80.3/22400

Xmol of H2 = 3.58×10¯³ mole

Therefore, 3.58×10¯³ mole of Hydrogen gas was released.

Now, we can determine the mole of the unknown metal as follow:

The balanced equation for the reaction is given below:

X + 2HCl —> XCl2 + H2

From the balanced equation above,

1 mole of the unknown metal reacted to produce 1 mole of H2.

Therefore, 3.58×10¯³ mole of the unknown metal will also react to produce 3.58×10¯³ mole of H2.

Therefore, the number of mole of the unknown compound is 3.58×10¯³ mole.

5 0
3 years ago
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