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3 years ago

What is 7.09x10 to the 5 minus 6.10xto the 4 in scientific notation

1 answer:

4 0

When subtracting scientific notation, you first need to make the exponents the same. The standard is to make the smaller exponent the same as the higher one.
Let's take a look at your given:

$7.09 x 10^{5}$
$6.10 x 10^{4}$ This is smaller.

Every time you add to the exponent, you need to move the decimal. So to make it the same, you need to add 1 to the 4 to get 5. Then move the decimal to the left. Your new notation would be:
$.610 x 10^{5}$

Then you just line up the decimal as you would before subtracting and just copy the base and exponent.
$7.09 x 10^{5}$
- $.61 x 10^{5}$
=
$6.48 x 10^{5}$

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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

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3 years ago

The azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib

stich3 [128]

Explanation:

Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.

The azide ion that is $N^{-}_3$ , is a symmetrical ion, all of whose contributing structures have formal charges.

Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.

Contributing structures of azide ion are drawn in the image attached.

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3 years ago

Please help!! I don’t think I did this right.?

harkovskaia [24]

Answer:

It was done correctly, Your answer is completely correct.

Explanation:

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2 years ago

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M=7M(H₂O)

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3 years ago

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maks197457 [2]

Answer:

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