4. numerical response\ncalculate the ph, to the nearest tenth, of a solution with a hydrogen - ion…

4. numerical response\ncalculate the ph, to the nearest tenth, of a solution with a hydrogen - ion concentration of 6.2×10⁻³ mol/l. (record your answer in the numerical response answer section)\nthe ph of a solution can be determined using the formula ph = -logh⁺\nwhere the concentration of hydrogen ions, h⁺, in a solution is measured in moles per litre.\nuse the following information to answer the next question\n7. almed borrows $27000 from his parents to buy a car. his parents charge him interest at a rate of 2%/a compounded quarterly. when he pays off the loan, he has to pay $28932.18 interest. what was the length of the loan?\na. 2 years\nb. 5 years\nc. 7 years\nd. 10 years\ncompound interest can be calculated using a = p(1 + i)ⁿ\nwhere,\na represents the final amount\np represents the initial principle amount\ni represents the interest rate per compounding period\nn represents the number of compounding periods

4. numerical response\ncalculate the ph, to the nearest tenth, of a solution with a hydrogen - ion concentration of 6.2×10⁻³ mol/l. (record your answer in the numerical response answer section)\nthe ph of a solution can be determined using the formula ph = -logh⁺\nwhere the concentration of hydrogen ions, h⁺, in a solution is measured in moles per litre.\nuse the following information to answer the next question\n7. almed borrows $27000 from his parents to buy a car. his parents charge him interest at a rate of 2%/a compounded quarterly. when he pays off the loan, he has to pay $28932.18 interest. what was the length of the loan?\na. 2 years\nb. 5 years\nc. 7 years\nd. 10 years\ncompound interest can be calculated using a = p(1 + i)ⁿ\nwhere,\na represents the final amount\np represents the initial principle amount\ni represents the interest rate per compounding period\nn represents the number of compounding periods

Answer

Explanation:

Step1: Identify the formula for pH

The formula for pH is $pH = -\log[H^+]$. Given $[H^+]=6.2\times 10^{-3}\text{ mol/L}$.

Step2: Substitute the value into the formula

$pH=-\log(6.2\times 10^{-3})$. Using the property of logarithms $\log(ab)=\log a+\log b$, we have $pH = -(\log6.2+\log(10^{-3}))$. Since $\log(10^{-3})=- 3$ and $\log6.2\approx0.792$, then $pH=- (0.792 - 3)$.

Step3: Calculate the pH value

$pH=-0.792 + 3=2.208\approx2.2$.

Answer:

$2.2$