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3 years ago

Calculate the pH of a solution that has [H3O+]=4.3 x 10^-5 M

Chemistry

1 answer:

Neko [114]3 years ago

6 0

Answer:

pH = 4.4

Explanation:

pH = -log[H₃O⁺]

= -log(4.3 × 10⁻⁵)

pH = 4.37 = 4.4

I'm actually learning this in my chemistry class right now lol. Hope this helps though. :)

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A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

RoseWind [281]

Answer:

$C_3=0.125M$

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

$V_1C_1=V_2C_2$

So we solve for C2:

$C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M$

2. Now, since 111 mL of water is added, we compute the final volume, V3:

$V_3=139+111=250mL$

So, the final concentration of the 139 mL portion is:

$C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M$

Best regards!

8 0

2 years ago

How many chlorine ions are required to bond with one aluminum ion

Monica [59]

The answer is Three
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4 0

2 years ago

For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(

OLga [1]

Answer:

1. 10.8 g of NO

2. N₂ is the limting reagent

3. 3.2 g of O₂ does not react

Explanation:

We determine the reaction: N₂(g) + O₂(g) → 2NO(g)

We need to determine the limiting reactant, but first we need the moles of each:

5.04 g / 29 g/mol = 0.180 moles N₂

8.98 g / 32 g/mol = 0.280 moles O₂

Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂

Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO

Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 = 0.360 moles NO

If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g

As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.

As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.

0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.

8 0

2 years ago

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sashaice [31]

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3 years ago

9. Formation of Coloured Ions

denpristay [2]

Um a rainbow wand others colors ! Hoped I help!

4 0

3 years ago