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Rasek [7]
3 years ago
11

What is the percent by volume of isopropyl alcohol in a solution that contains 45 ml of isopropyl alcohol in 1.4 L of water?

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Answer:

3.11%

Explanation:

We express a solution's volume by volume percent concentration, % v/v,

Take the ratio of the isopropyl alcohol (IPA) volume to the total volume of the solution, which is 1400 mL of water+ 45 mL of IPA,

and multiply by 100 to get the percentage:    45/(1400+45) *100 = 0.0311*100 =  3.11%

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Suppose you find a rock that contains some potassium-40 (half-life of 1.3 billion years). you measure the amount and determine t
Alborosie
Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time

5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t

And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
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It went through 3 half-lifes
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3 years ago
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Explanation:

8 0
3 years ago
Read 2 more answers
You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli
maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

M_1V_1=M_2V_2

where, M_1 is initial molarity and  M_2 is the molarity after dilution. Similarly,  V_1 is the volume before dilution and  V_2 is the volume after dilution.

Let's plug in the values in the equation:

1.25M(363mL)=0.50M(V_2)

V_2=\frac{1.25M(363mL)}{0.50M}

V_2=907.5mL

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

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3 years ago
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The nucleons incorporate neutrons and protons. All the positive charge of an atom is found in the nucleus and arises from the protons, the neutrons are neutrally-charged, and the electrons are the negatively charged particles found outside of the nucleus.  


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