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3 years ago

15

a few drops of perfume were spilt on the floor a few minutes later ghe perfume could be smelt a few meters away what 2 processes

had taken

1 answer:

MArishka [77]3 years ago

6 0

Answer:

evaporation and diffusion

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The metalloid that has three valence electrons is?

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The metalloid that has three valence electrons is Boron~

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An increase in aurora activity on Earth could be related to what activity on the Sun?

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Answer:

D.

Explanation:

INCREASE OF SOLAR WINDS When the sun is more active

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2 years ago

which of these solutions has the lowest freezing point? 0.25 m nacl 0.5 m nacl 1.0 m nacl 1.5 m nacl 2.0 m nacl

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The more particles (ions or molecules) that you can put into solution, the lower the freezing point.

the answer is E. 2.0 M nacl

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3 years ago

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A solution contains 1 LaTeX: \times\:×10−4 M OH– ions. Calculate the solution pH value, and determine if the solution is acidic,

oksano4ka [1.4K]

<u>Answer:</u> The pH value of the solution is 10 and the solution is basic in nature.

<u>Explanation:</u>

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=1\times 10^{-4}M$

Putting values in above equation, we get:

$pOH=-\log(1\times 10^{-4})\\\\pOH=4$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4=10$

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

The pH range of acidic solution is 0 to 6.9
The pH range of basic solution is 7.1 to 14
The pH of neutral solution is 7.

As, the pH of the solution is 10 and is lying in the range of basic solution, so the solution is basic in nature.

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3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

<u>Answer:</u> The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

<u>For chromium (III) acetate:</u>

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

<u>For chromium (III) nitrate:</u>

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

<u>For chromium cation:</u>

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

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3 years ago