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charle [14.2K]
3 years ago
6

Calculate the pH of a solution that has a [OH−] of 2.6 × 10^−6 M.

Chemistry
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:

A. 8.4

Explanation:

[OH⁻] = 2.6 × 10⁻⁶               Take the negative log of each side

-log[OH⁻] = pOH = 5.59     Apply the pH/pOH relation

pH + pOH = 14.00               Insert the value of pOH

pH + 5.59 = 14.00               Subtract 5.59 from each side

pH = 14.00 -5.59 = 8.41

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2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

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3 years ago
An aqueous solution containing 35.5 g of an unknown molecular (non-electrolyte) compound in 151.0 g of water was found to have a
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ccccccccccccccccccccccccccccccccccc

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4 years ago
List down the observations made by Rutherford during the gold foil experiment.
Travka [436]

Answer:

1) The volume occupied by an atom is composed of mainly empty space

2) Atoms have a very small, relatively dense, central nucleus that is positively charged

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Explanation:

The selection of gold for the gold foil experiment was due to its ability to be rolled into extremely thin sheets such that it was expected for alpha particle to perforate or pass through the foil.

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3 years ago
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