All categories

3 years ago

What is the [ H3O+] of a solution with pH = (8.50x10^0)? enter your answer in scientific notation and with three sig figs.

Chemistry

2 answers:

blagie [28]3 years ago

4 0

Answer:

3.17 x 10^-9

Explanation:

log 3.17 =.5 log 10^-9 =-9 .5+-9=-8.5 =pH 8.5

erastovalidia [21]3 years ago

3 0

Answer:

[H₃O⁺] = 3.162 × 10⁻⁹ moles / liter

Explanation:

The pH is calculated with the formula:

pH = -lg[H₃O⁺]

From here the concentration of hydronium ion (H₃O⁺) will be:

[H₃O⁺] = $10^{-pH}$

[H₃O⁺] = $10^{-8.5}$

[H₃O⁺] = 3.162 × 10⁻⁹ moles / liter

You might be interested in

NEED ASAP

notka56 [123]

Answer: B

Refer:

Because water molecules are polar, they interact with the sodium and chloride ions. In general, polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes. This concept is often expressed as “Like dissolves like.

5 0

3 years ago

How the relative density of a substance is related to the density calculate the density of iron if its relative density is 2 and

il63 [147K]

Answer:

How the relative density of a substance is related to the density calculate the density of iron if its relative density is 2 and a density of water is 2gcm -3

R.d= relative density of substance/ relative density of water

R.d= 2/2

R.d= 1gcm-3

Explanation:

5 0

3 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

3 years ago

The molar solubility of pbi2 is 1.5 103 m.

Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

8 0

3 years ago

When .50 mL of 2.0M Li3PO4 is added to 50 mL of 1.0 M LiH2PO4 what is the resulting concentration of Li+?

jeka57 [31]

Hence the resulting concentration of Li+ is 1M. Hope it helps.

8 0

3 years ago