How does adding salt to water affect the freezing point

How does your Skeletal and Muscular Systems work together?

elena-s [515]

Answer:

Answer would be A, Muscles move your bones

Explanation:

Without the Muscular System, your bones would not be connected properly

3 0

3 years ago

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Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?

trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution = 2.5L

Mass of Calcium phosphate = 600g

Unknown:

Concentration = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

Formula of calcium phosphate = Ca₃PO₄

molar mass = 3(40) + 31 + 4(16) = 215g/mol

Number of moles of Ca₃PO₄ = $\frac{600}{215}$ = 2.79moles

Now;

Concentration = $\frac{Number of moles }{volume }$

Concentration = $\frac{2.79}{2.5}$ = 1.12M

7 0

3 years ago

Water is called compound why?

qaws [65]

Answer: because it consists of more than one element, which are hydrogen and oxygen in a covalent bond.

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3 years ago

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When you combine chocolate powder and milk, the result is best describe as?

NeX [460]

Answer:

The answer to your question would be substance, but chocolate power mixed into milk would be more of a suspension.

Explanation:

Neither chocolate powder nor milk are elements. They are both complex molecules. Their mixture will not result in the formation of a compound since no chemical reaction will take place.

The molecules of the chocolate powder will simply intermingle with the fatty molecules of the milk to form the substance.

When thoroughly mixed the solution will become homogeneous so there will be no lumps of chocolate power visible. But after time, the chocolate will become visible at the bottom of the clear container in which we asked you to prepare the mixture.

HOPE THIS HELPS :)

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3 years ago

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