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iVinArrow [24]
2 years ago
8

What happens to the total mass of a substance undergoing a physical change

Chemistry
2 answers:
yKpoI14uk [10]2 years ago
7 0
That depends. What kind of change are you talking about? But

Mass<span> through chemical </span>change<span> stays the same as well. Example: burning paper, the ash left behind is not all of the </span>mass<span> of the reactants, Carbon dioxide, and other </span>substances<span> also makeup</span>mass<span> but just is not seen
</span>
Lady_Fox [76]2 years ago
6 0

If your answer choices are:

It stays the same.

It doubles.

It triples.

It is reduced.

The correct answer is "It stays the same."

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a 0.5678 of KHP required 26.64cm³ of NaOH to complete neutralization.calculate the molarity of the NaOH solution​
vladimir2022 [97]

Answer:

Explanation:

0.5678 G        X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL               1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE  KHC8H4O4=0.00278 MOL NaOH

0.00278 MOL NaOH/26.26ml=0.106 molar

4 0
2 years ago
Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
stira [4]

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

8 0
2 years ago
Material A has a small latent heat of fusion. Material B has a large heat of fusion. Which of the following statements is true?
Olin [163]
The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
7 0
2 years ago
Read 2 more answers
Both black and white road surfaces radiate energy. at midnight on a starry night the warmer road surface is the
luda_lava [24]
The black road because during the day it absorbed more radiation than the with one
5 0
1 year ago
When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi
jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be  PO43- < NO3- < Na+

4 0
2 years ago
Read 2 more answers
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