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xxTIMURxx [149]

2 years ago

12

5. Rubbing alcohol is a commonly used disinfectant and has a cooling effect when applied to the skin. The active ingredient in r

ubbing alcohol is isopropanol. In drugstores, the most common concentration of rubbing alcohol sold contains 70% (vol/vol) isopropanol in water. Assuming the rubbing alcohol manufacturer uses a 100% isopropanol solution, what volume of pure isopropanol is required to produce a 200-mL bottle of rubbing alcohol

1 answer:

WITCHER [35]2 years ago

8 0

Answer:

Explanation:

70% (vol/vol) means

cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.

if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.

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The elements chlorine and iodine have similar chemical properties because they _____. Group of answer choices have the same numb

worty [1.4K]

Answer:

Have the same number of electrons in their outer energy levels

Explanation:

Elements in the same group have similar chemical properties because they have the same number of valence electron(s) in their outermost shell.

Chlorine and Iodine have similar chemical properties because they have the same number of valence electron in their outermost shell. This can be seen from their electronic configuration as shown below:

Cl (17) => 1s² 2s²2p⁶ 3s²3p⁵

I (53) => [Kr] 4d¹⁰ 5s²5p⁵

From the above illustration:

Outer shell of Cl (3s²3p⁵) = 2 + 5 = 7 electrons

Outer shell of I (5s²5p⁵) = 2 + 5 = 7 electrons

Since they have the same number of valence electrons, therefore, they will have similar chemical properties.

7 0

1 year ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

1 year ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

Why are metallic crystals malleable and ductile? A. The electrons are free floating, allowing them to move with the atoms when t

jok3333 [9.3K]

Answer:

D. The atoms are arranged with alternating positive and negative charges. When struck, the lattice shifts putting positives against positives and negatives against negatives.

Explanation:

Metallic crystals takes their properties as a result of metallic bonds in between the atoms.

Metallic bond is actually the attraction between the positive nuclei of all the closely packed atoms in the lattice and the electron cloud jointly formed by all the atoms by losing their outermost shell electrons this is by virtue of their low ionization energy.

Physical properties of metals such as malleability, ductility, electrical conductivity, etc can be accounted for by metallic bonds.

7 0

2 years ago

How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K

Olenka [21]

I’m not sure I need help with this question

6 0

1 year ago