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3 years ago

12

Which chemical do you think would best maintain the pH of the pool water?

2 answers:

Pepsi [2]3 years ago

7 0

The answer is Bleach.

Explanation:

Hey, I might not be the best in chemistry but I do know the answer to this question!

malfutka [58]3 years ago

6 0

Answer: Bleach

Explanation:

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Explain why it is harder to remove an inner shell electrons than a valence electron from an atom

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<span> It makes sense that an inner shell electron would be tougher to remove than a valence electron because the inner shell electron is closer to the positive nucleus of the atom. Seeing as an electron caries a negative charge it would be too attracted to the positive core to leave readily. Also, the inner shell electrons are constantly repelling electrons outside of it's energy level (however the reason these electrons outside innershell energy levels don't simply fly away is the charge of the positive core overcomes the smaller charges of the comparably negligible inner shell electrons, but that repulsion is still there so keep that in mind) </span>

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Carboxyhemoglobin may be formed when a person inhales air that contains

elena-14-01-66 [18.8K]

Answer is none of the above

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3 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%

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4 years ago

How many “dots” would be around the central atom in the Lewis structure of S?

agasfer [191]

B) 6
one above one below and 2 on the left and right sides

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3 years ago

Read 2 more answers