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SashulF [63]
3 years ago
9

in an experiment magnesium ribbon was heated in air the product formed was found heavier than original ribbon.potassium manganat

e(vii)was heated im air and product formed was found lighter.explain different observations made​
Chemistry
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

1: due to difference in reactivity and melting point

2 during the reaction both combine with oxygen in this case magnesium remain solid but some potassium vaporise

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Which statement about silver Ag metal is least reliable
Anna007 [38]
<span>The choices for this question can be found elsewhere and as follows:

Silver is a transition metal with electrons in the d shell.
Silver atoms have 47 protons in their nuclei.
Silver atoms commonly form ions with a +1 charge.

I think the correct answer is the last option. The least reliable information would be that s</span>ilver atoms commonly form ions with a +1 charge.
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3 years ago
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Draw a diagram using H+ signs and OH- symbols of a solution that is acidic. Hint: Use the
Dimas [21]

More positive ions shows acidic whereas more negative ions indicates basic solution.

<h3>Which charge show acidic solution?</h3>

That side which has more positive charges is considered as acidic solution while on the other hand, that region where negative charges are present in large number as compared to positive charges is considered as basic or alkaline solution.

So we can conclude that more positive ions shows acidic whereas more negative ions indicates basic solution.

Learn more about charge here: brainly.com/question/25923373

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2 years ago
18-19. If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the a
miss Akunina [59]

Answer:

As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Explanation:

Absorbance of light by a reagent of concentration c, is given as

A = εcl

A = Absorbance

ε = molar absorptivity

c = concentration of reagent.

l = length of light path or length of the solution the light passes through.

So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.

But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.

Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Hope this Helps!!!

5 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
Which is the stronger acid in each of the following pairs? Explain your reasoning.
Makovka662 [10]

Answer:p-hydroxybenzaldehyde is stronger acid to phenol

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The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95. 

The pKa value of o-fluorophenol is 8.7, while that of the p-fluorophenol is 9.9. It's obvious that the inductive effect is more dominant at ortho-position, which results in a more acidic nature

The pKa is the pH value at which a chemical species will accept or donate a proton. The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.

6 0
3 years ago
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