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10 months ago

What are the hydronium and hydroxide ion concentrations in a solution whose ph is 6.52?

1 answer:

3 0

A solution with a pH of 6.52 has a hydronium ion concentration of 3.02x10-7 mol/L and a hydroxide ion concentration of 3.31x10-8 mol/L.

The hydronium ion concentration of a solution can be calculated from pH by using $10^{-pH}$ . For a pH of 6.52, hydronium ion concentration is 3.02x10-7 mol/L.

The concentration of hydroxide ions can be determined by identifying the value of pOH. The sum of pOH and pH is equal to 14, which is based on the negative logarithm of the ion-product constant of water. At a pH of 6.52, pOH is equal to 7.48.

The relationship between pOH and hydroxide ion concentration is the same as the relationship between pH and hydronium ion concentration. With this, the hydroxide ion concentration at pOH of 7.48 is $10^{-7.48}$ or 3.31x10-8 mol/L.

For more information regarding pH and pOH, please refer to the link brainly.com/question/13557815.

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Which set of numbers gives the correct possible values of l for n = 2

Paha777 [63]

HEY THERE!!

For a given principal quantum number or n, the corresponding angular quantum number or l is equivalent to a range between 0 and (n-1) .

This means that the angular quantum number for a principal quantum number of 2 is equivalent to:

l = 0 -> (n-1) = 0 -> (2-1) = 0 -> 1

So the answer is 0, 1

HOPE IT HELPED YOU.

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3 years ago

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When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro

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Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.

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3 years ago

We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to

andreyandreev [35.5K]

Answer: The correct answer is Option 5.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base.

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.

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Answer:

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