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3 years ago

A block of magnesium has a mass of 14.3 g and a volume of 8.46 cm3. What is the density of

Chemistry

1 answer:

sveta [45]3 years ago

7 0

Density is mass over volume, so:
14.3/8.46≈ 1.6903 g/cm^3

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Read the scientific question below.

svetlana [45]

If we fertilize a plant, then its height increases fast. Always use if then format

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3 years ago

A mass of 100.0 g of NaCl is added to 100.0 mL of water, but not all of it dissolves. A mass of 59.5 grams of NaCl solid remains

Taya2010 [7]

Answer:

The molarity of the dissolved NaCl is 6.93 M

Explanation:

Step 1: Data given

Mass of NaCl = 100.0 grams

Volume of water = 100.0 mL = 0.1 L

Remaining mass NaCl = 59.5 grams

Molar mass NaCl= 58.44 g/mol

Step 2: Calculate the dissolved mass of NaCl

100 - 59. 5 = 40.5 grams

Step 3: Calculate moles

Moles NaCl = 40.5 grams / 58.44 g/mol

Moles NaCl = 0.693 moles

Step 4: Calculate molarity

Molarity = moles / volume

Molarity dissolved NaCl = 0.693 moles / 0.1 L

Molarity dissolved NaCl = 6.93 M

The molarity of the dissolved NaCl is 6.93 M

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3 years ago

Read 2 more answers

On melting of ice , there is a decrease in volume instead of increase . Why ?

Oliga [24]

Because of the crystal structure of the ice, ice has lower density than liquid water. So the volume of the ice of same mass is greater than water. When melting, the volume will decrease.

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3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago

An acid

tekilochka [14]

The answer is A. Has a low pH in solution.

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3 years ago