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3 years ago

7

At 25 °c, what is the hydroxide ion concentration, [oh–], in an aqueous solution with a hydrogen ion concentration of [h ] = 3.0

× 10–5 m?

1 answer:

chubhunter [2.5K]3 years ago

5 0

As,

Water has a pkw=14

so it can be represented as,

[H+] [OH-] = 1*10^-14

If [H+] = 3*10^-5M

[OH-] = (1*10^-14) / ( 3*10^-5)

[OH-] = 3.3*10^-9 M

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Is wine a homogenous mixure

Fiesta28 [93]

Tap water and rain water are both homogeneous, even though they may have different levels of dissolved minerals and gases. A bottle of alcohol is a man-made homogeneous mixture, from a fine Italian wine to a glass of Scotch whisky. In the human body, blood plasma is an example of a homogeneous mixture.

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3 years ago

Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

<u>Answer:</u>

<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

<u>For A:</u> (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

<u>For B:</u> (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

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3 years ago

The article explains that elephants sometimes communicate in low ultrasound. Based on what you learned above, would infrasound w

Likurg_2 [28]

I have not read the article, and so I don't know in what context this is referring to. But I do know that <span>Infra-sound is made up of a really low frequency sound, beyond the range of hearing by humans. </span>Helpful? :)

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3 years ago

Examination of the first few successive ionization energies for a given element usually reveals a large jump between two ionizat

gladu [14]

Answer:

See Explanation

Explanation:

Ionization energy refers to the energy required to remove an electron from an atom. Metals have lower ionization energy than non metals since ionization energy increases across a period.

One thing that we must have in mind is that it takes much more energy to remove an electron from an inner filled shell than it takes to remove an electron from an outermost incompletely filled shell.

Now let us consider the case of magnesium which has two outermost electrons. Between IE2 and IE3 we have now moved to an inner filled shell(IE3 refers to removal of electrons from the inner second shell) and a lot of energy is required to remove an electron from this inner filled shell, hence the jump.

For aluminium having three outermost electrons, there is a jump between IE3 and IE4 because IE4 deals with electron removal from a second inner filled shell and a lot of energy is involved in the process hence the jump.

Hence a jump occurs each time electrons are removed from an inner filled shell.

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AlladinOne [14]

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3 years ago

Read 2 more answers