Can anyone help me with this? Even just a few answers would be helpful. I'm struggling a little bit

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

erastovalidia [21]

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

3 0

3 years ago

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PLEASE HELP I BEG

Gwar [14]

Answer:

1.Velocity and pitch

2.The unit of vibration is hertz. The number of oscillations of an oscillating body per second is known as frequency. Complete step by step answer: The time taken by a vibrating body to complete one vibration is time period.

3.Frequencies above 20,000 Hz are known as ultrasound.

5 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

*GIVING BRAINLIEST* NEED ANSWER ASAPPP

poizon [28]

Same Question here answered by me with explanation check the link below for your answer.

brainly.com/question/24944271?

8 0

2 years ago

What hypothesis led to the discovery of the proton?

Lana71 [14]

When a neutral hydrogen atom loses an electron, a positively-charged particle should remain.

4 0

3 years ago

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