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nadya68 [22]
3 years ago
14

Devon needs 90mL of a 10% acid solution for a science experiment. He has available a 5% solution and a 20% solution. How many mi

lliliters of the 5% solution and how many milliliters of the 20% solution should he mix to make the 10% solution?
Chemistry
1 answer:
Stels [109]3 years ago
6 0

Answer:

Devon needs add 60 mL of 5% acid solution and 30 mL of 20% acid solution

Explanation:

It know acid and water concentration for every solution. 2 equation are proposed because there are two unknowns. V1= volume 5%acid and V2 =volume 20% acid

<em>Acid   </em>V1(0.05)+V2(0.2)=90mL(0.1)

<em>Water   </em>V1(0.95)+V2(0.8)=90mL(0.9)

Clearing V1 from acid equation having to: V1=180-4V2

Replacing V1 in water equation having to: (180-4V2)(095)+V2(0.8)=81

Clearing V2 = 30mL

Replacing in water equation V1(0.05)+30(0.2)=9  So, V1=60mL

I hope it helps

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Explanation:

<em>Ionic compounds</em> are formed by the electrostatic attraction of cations and anions.

Cations, positive ions, are formed when atoms lose electrons, and anions, negative ions, are formed when atoms gain electrons.

When two different atoms have similar atraction for electrons (electronegativity) they will not donate to nor catch electrons from each other, so cations and anions will not be formed. Instead, the atoms would prefer to share electrons forming covalent bonds to complete their outermost shell (octet rule).

Then, in order to form ionic compounds the electronegativities have to substantially different. This situation does not happen between two nonmetal elements, which nitrogen and sulfur are. Then, you can predict safely that nitrogen and sulfur will not form an ionic compound.

Ionic compounds, then require the electronegativity difference that exist between some metals and nonmetals. Being magnesium an alkaline earth metal, its electronegativity is very low. On the other hand, fluorine the first element of the group 17, has the highest electronegativity of all the elements.Thus magnesium and fluorine will have enough electronegativity difference to justify the exchange of electrons, forming ions and, consequently, ionic compounds.

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4. Option C. Pentane.

5. Option D.

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4. Hydrocarbons are compound containing carbon and hydrogen only. Hydrocarbons are said to be saturated when they contain only carbon to carbon single bond. All alkanes are saturated hydrocarbon.

The correct answer is pentane.

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What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250
Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons \text{H}^{+} than hydroxide ions \text{OH}^{-}. The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that x \; \text{L} of the 0.307 \text{mol} \cdot \text{dm}^{-3} sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, x is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain 0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x moles of acetic acid and 0.307 \; x moles of acetate ions.

Let \text{HAc} denotes an acetic acid molecule and \text{Ac}^{-} denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of 10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}. There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by 10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3} would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}

By definition:

\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)

The question states that

\text{K}_{a} = 1.75 \times 10^{-5}

such that

10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241

Thus it takes 0.0241 \; \text{L} of sodium hydroxide to produce this buffer solution.

6 0
3 years ago
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