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2 years ago

A substance with a high pOH would likely have which of the following?

Chemistry

2 answers:

Anna [14]2 years ago

8 0

Answer: a low $OH^-$ and low pH.

Explanation:

pH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pOH=log\frac {1}{OH^-}$

Thus as pOH and $OH^-$ are inversely related, a solution having higher pOH will have less amount of $OH^-$ concentration. And a solution having more pOH will have less pH.

Thus a substance with a high pOH would likely have low $OH^-$ concentration and low pH.

Zinaida [17]2 years ago

4 0

Answer:

A low [OH– ] and a high pH

Explanation:

I just took the test

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I need some help with chemistry. Let me know if you are good at it!

tekilochka [14]

Answer:

I think im good at it i have an A in the class

Explanation:

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2 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

Calculate the bond energy per mole for breaking all the bonds in methane, CH4. Express your answer to four significant figures a

lesantik [10]

The energy that is essential to break one C-H bond is 414 kJ/mol. Since, there are four C-H bonds in CH4, the energy Δ HCH4 for breaking all the bonds is calculated as Δ HCH4 = 4 x bond energy of C-H bond. By multiplying the 4 with the 414 kJ/mol you can get the answer of 1656 kJ/mol CH4 molecules.

4 0

2 years ago

Read 2 more answers

What are the 5 ways to identify if a chemical reaction happened?

vlada-n [284]

Answer:

a new substance is formed, if heat is formed, color change, the evolution of gas, and bubbles or fizzing

7 0

3 years ago

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter?

NARA [144]

Answer:

ΔH = $q_{p}$

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

$q_{com}$ = $q_{p}$

where

$q_{p}$ = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

$q_{com}$ = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is $q_{p}$

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = $q_{p}$ + 0

==> ΔH = $q_{p}$

5 0

3 years ago