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inessss [21]
3 years ago
7

a bottle of rubbing alcohol is labeled 70%(v/v) how many milliliters of isopropyl alcohol are in a 400ml bottle of this solution

?
Chemistry
1 answer:
xz_007 [3.2K]3 years ago
3 0
 The number of   milliliters  of  isopropyl alcohol that are  in   a 400 ml bottle  of this solution  is 280 Ml

   calculation

  % V/V =  volume of  solute/ volume  of  solution  x 100
 Let the volume  of solute  be =  y  ml
 the volume  of  solution =  400 ml
convert 70%  to fraction = 70/100

 by  substituting  the value  in the equation 

 = y  ml/  400 ml =  70/100

by cross  multiplication

100y =400 x70
100y=  28000
y= 280 ml
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For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:
KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0
3 years ago
A small cube of iron and a large flat sheet of iron contain the same volume. Which one will completely rust first? Explain why.
Hunter-Best [27]

The flat sheet will completely rust before the iron cube. Since they both have the same volume, the flat sheet has more surface area than the small cube. This means more particles are exposed on the flat sheet that can react in a chemical reaction.

4 0
3 years ago
Read 2 more answers
How many moles of methane (CH4) are in 7.31 x 10^25 molecules
Ivan

molar mass of methane CH4

= C + 4 H  

= 12.0 + 4 x 1.008

= 12.0 +  4.032

= 16.042g/mol

7.31 x 10^25 molecules x <u>             1 mole  CH4    </u>  = 121.43 moles

                                       6.02 x 10^23 CH4 molecules

121.43 moles CH4 are present.

       


6 0
3 years ago
Read 2 more answers
Which is the balanced equation for magnesium added to copper(i) chloride yielding copper and magnesium chloride?
erastova [34]
Hkjghltuilkjh9uopiuk,l;ui
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3 years ago
When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

m= molality = \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579

8.30^0C=1\times K_f\times 0.579

K_f=14.3^0C/m

Let Mass of solute (KBr) = x g

8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0
2 years ago
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