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postnew [5]
2 years ago
12

Amol has divided some substances into two groups as shown in the table below. He wants to further separate all the substances in

Group 1 again into exactly two groups. Which of the following criteria will allow him to do so?
A. liquids and solids  

B. mixtures and pure substances

C. naturally occurring and man-made substances.

D. substances completely soluble and insoluble in water.​

Chemistry
2 answers:
Nonamiya [84]2 years ago
6 0

The criteria that will allow Amol separate group 1 substances into two groups are mixtures and pure substances (option B).

<h3>What are mixtures and pure substances?</h3>

Mixtures are substances that are made up of two or more substances that can be easily separated.

Pure substances, on the other hand, are elements and compounds that cannot be separated by chemical means.

In group 1 of the above table, the substances can be classified as follows:

  • Soil - mixture
  • Tomato ketchup - mixture
  • Smoke - compound
  • Milk - mixture
  • Hair shampoo - mixture

Therefore, the criteria that will allow Amol separate group 1 substances into two groups are mixtures and pure substances.

Learn more about mixture at: brainly.com/question/24898889

#SPJ1

Degger [83]2 years ago
5 0

Answer:

Explanation:

b is the right answer

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What is the acceleration of a 50kg object pushed with a force of 500 Newton’s
olasank [31]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m} \\

f is the force

m is the mass

From the question we have

a =  \frac{500}{50}  =  \frac{50}{5}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

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3 years ago
The bond order denotes:_____.1. the order of filling of bonding vs. antibonding orbitals. 2. the pattern in which Lewis dot stru
tiny-mole [99]

Answer:

an estimate of the strength of a bond.

Explanation:

The bond order is given as;

1/2(number of bonding electrons - number of anti bonding electrons)

The bond order tells us about the strength of bond. As the bond order increases, so does the strength of the bond because atoms involved in bonding come closer to each other as the bond length decreases.

Hence, bond order is an index of bond strength. Triple bonds are stronger than double bonds which are stronger than single bonds.

6 0
3 years ago
A zinc block with a mass of 230 g is given 1320 J of energy. What is the change in
Mariana [72]

Answer:

14.7°C

Explanation:

Q = m·ΔT·c

ΔT = \frac{Q}{m*c}

ΔT =\frac{1320 J}{230 g* 0.39 J/gC}

     = 1320 J / ((230 g) * (.39 J/g°C)

ΔT = 14.7 °C      

4 0
3 years ago
50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the
scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

H_2NC_2H_5CO^-_2  +  H^+  ------>  H_3}^+NC_2H_5CO^-_2

1 mole of  alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL  of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via  the following process as shown below;

[H_3}^+NC_2H_5CO^-_2] = \frac{initial moles of alaninate}{total volume}

[H_3}^+NC_2H_5CO^-_2] = \frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}

[H_3}^+NC_2H_5CO^-_2] = \frac{8}{100mL}

[H_3}^+NC_2H_5CO^-_2] = 0.08 M

Alanine serves as an intermediary form, however the concentration of H^+ and the pH can be determined as follows;

[H^+] = \sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{  K_{a1}{[H_3}^+NC_2H_5CO^-_2]  } }

[H^+] = \sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})}  {(10^{-pK_{a1}})+(0.08)} }

[H^+] = \sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})}  {(10^{-2.344})+(0.08)} }

[H^+] =  7.63*10^{-7}M

pH = - log [H^+]

pH = -log[7.63*10^{-7}]

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

H_2NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO^-_2

H^+_3NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO_2H

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

[H^+_3NC_2H_5CO_2H] = \frac{initial moles of alaninate}{total volume}

[H^+_3NC_2H_5CO_2H] = \frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}

[H^+_3NC_2H_5CO_2H] = \frac{8}{150}

[H^+_3NC_2H_5CO_2H] = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

H^+_3NC_2H_5CO_2H        ⇄        H^+_3NC_2H_5CO^-_2    +  H^+

(0.053 - x)                                  x                             x

K_{a1} = \frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}

10^{-PK_{a1}} = \frac{x*x}{(0.053-x)}

10^{-2.344} =\frac{x^2}{(0.053-x)}

0.00453 = \frac{x^2}{(0.053-x)}

0.00453(0.053-x) =x^2

x^2+0.00453x-(2.4009*10^{-4})

Using quadratic equation formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

we have:

\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)} OR \frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}

= 0.0134                    OR                -0.0179

So; we go by the positive integer which says

x = 0.0134

So [H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M

pH = -log[H^+]

pH = -log[0.0134]

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0
3 years ago
Would Nitrogen Dioxide be ionic or covalent
tatiyna

Explanation:

The nitrogen dioxide is a covalent compound where one nitrogen is the central atom which is bonded to two oxygen atoms, where one oxygen atom is bonded by a single bond and other oxygen atom by a double bond.

4 0
3 years ago
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