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Zina [86]
3 years ago
11

You need 300 ml of a 75% alcohol solution. on hand, you have a 10% alcohol mixture and a 85% alcohol mixture. how much of each m

ixture will you need to obtain the desired solution?
Chemistry
1 answer:
posledela3 years ago
3 0

Well, to find how much of the Alcohol Solution you will need, you will need to do;

300 (divided by) 10, which will give you a answer of 30. Then you will do 30 (times) by 7, which will give you 210, which will then give you 70% of your 75%. Then, to find the other 5%, you need to do what the answer was to 10%, and divide it by 2, so in this case, the answer will be 15. So now to get your full 75%, you add your working out together for the 70% and the 5% which gives you 225.

Alcohol Solution = 225

Then, to find out what amount of the Alcohol Mixture you need, you will do;

300 (divided by) 10 which is 30, which will give you the answer to how much Alcohol Mixture you need.

Alcohol Mixture {1} = 10

To find out the second amount of how much of the other Alcohol Solution you will need, you need to;

Find 10% of 300, which is 30 (300 {divided by} 10), then you will times that answer (30), by 80 to find your 80%, which will be 240, then you will need to find your other 5%, so you will need to do 300 (divided by) half of the answer you got before you timed the answer by 8, so that would be 15 (added onto) 240, which will give you an overall answer of 255.

Alcohol Mixture {2} = 255

The ones in bold are your answers in order, hope they help! xo

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Create a 3-D model of Bohr's atom for lithium
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https://sciencing.com/make-3d-model-atom-5887341.html


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3 years ago
If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?
ICE Princess25 [194]

<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

t = time taken = 1.52 hrs

[A_o] = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

6 0
3 years ago
Identify the single displacement reaction. 2H 2 + O 2 ⟶ 2H 2O Al 2S 3 ⟶ 2Al + 3S Cl 2 + 2KBr ⟶ 2KCl + Br 2 C 4H 12 + 7O 2 ⟶ 6H 2
ozzi

Answer: Cl_2+2KBr\rightarrow 2KCl+Br_2

Explanation:

A single displacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.

Cl_2+2KBr\rightarrow 2KCl+Br_2

Synthesis reaction is defined as the reaction where substances combine in their elemental state to form a single compound.2H_2+O_2\rightarrow 2H_2O

Decomposition reaction is defined as the reaction where a single substance breaks down into two or more simpler substances.

Al_2S_3\rightarrow 2Al+3S

Combustion is a type of chemical reaction in which hydrocarbons burn in the presence of oxygen to form carbon dioxide and water along with heat.

C_4H_{12}+7O_2\rightarrow 6H_2O+4CO_2

6 0
3 years ago
What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?
tamaranim1 [39]
<h2>Answer:AlF_{3} </h2>

Explanation:

The chemical equation of the reaction that occurs when AlF_{3} reacts with O_{2} is

4AlF_{3}+3O_{2}→2Al_{2}O_{3}+6F_{2}

4 moles of AlF_{3} requires 3 moles of O_{2}.

1 mole of AlF_{3} requires \frac{3}{4} moles of O_{2}.

Given that we have 9 moles of AlF_{3}.

9 moles of AlF_{3} requires \frac{3}{4}\times 9=6.75 moles of O_{2}.

But we have 12 moles of O_{2}.

So,AlF_{3}  will be consumed first.

So,AlF_{3}  is the limiting reagent.

3 0
3 years ago
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