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Mrrafil [7]
2 years ago
15

b. Sugar and salt are both white, crystalline powders that dissolve in water. If you were given an unknown sample that contained

one or both of these solids, how could you determine what your unknown sample contained (without tasting it)? (2 points)
Chemistry
1 answer:
NikAS [45]2 years ago
5 0

Answer:

Explanation:

Sugar and Salt even thought they both dissolve in water they both dissolve in different ways. When salt dissolves in water, its individual types of ions are torn apart from each other, while Sugar molecules stay together when dissolved in water, and therefore the molecules remain the same when dissolved in water. This being said in science using your senses can be just as valuable as using calculations. In this case both Sugar and Salt taste differently. Sugar is sweet while Salt is salty. Therefore tasting the substance can be the easiest and most accurate way of determining the substance.

Read more at Answer.Ya.Guru – https://answer.ya.guru/questions/4803546-sugar-and-salt-are-both-white-crystalline-powders-that-dissolve.html

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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
2 years ago
Which statement best explains why the elements in Group 18 do not have electronegativity values?
Gnesinka [82]

Answer:

1) The elements have filled valence levels.

Explanation:

Since they have filled valence levels, they're stable and don't need to electrons to fill their valence shells since they're already full.

2) False, They do have electrons

3) False, He does have only one electron shell, but going down the periods, every next element have one more electron shell than a preceding one has.

4)False, they're actually the smallest atoms of their respective period

5 0
2 years ago
How many electrons are in Hydrogen?
KonstantinChe [14]
1. Hydrogen has 1 electron.
5 0
2 years ago
Read 2 more answers
A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium
babunello [35]
First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume 

                    = 0.15 m * 0.025 L

                   =  0.00375 moles

moles NaOH = molarity * volume 

                       = 0.15 m * 0.015 L

                       = 0.00225 moles 

after that we shoul get the total volume = 0.025L + 0.015L

                                                                   = 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume 

           = 0.00375 moles / 0.04 L

           = 0.09375 M

∴[NaOH] = moles NaOH / total volume 

                = 0.00225 moles / 0.04 L

                = 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka 

       = - ㏒ 1.9 x10^-5

      = 4.7 

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48 


6 0
2 years ago
Which of the following is accurate in describing the placement and classification of iodine?
fenix001 [56]
You can answer this question by only searching the element in the periodic table.

The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).

The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
4 0
2 years ago
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