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Margarita [4]
3 years ago
13

Why is it NOT possible for ONE particle (atom/ion/molecule) to be classified as a particular phase, or state, of matter?

Chemistry
1 answer:
Agata [3.3K]3 years ago
6 0

Answer:

because I said and I'm the boss

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Predict how many grams of KCI is produced from 40 grams of K?
Nutka1998 [239]

Explanation:

firstly find for the molar mass of kcl and molar mass of k

and then

molar mass of k = x

molar mass of kcl= 40

cross mutiply and then simplify you will get your answer

5 0
3 years ago
why does ocean water found near the poles often have a higher leaves of salinity than ocean water found near the equator
zaharov [31]

Answer:

Water near the poles often have higher salinity because Cold polar air cools the water and lowers its temperature, increasing its salinity. Fresh water freezes out of seawater to become sea ice, which also increases the salinity of the remaining water.

Understands! ♥

4 0
3 years ago
Would precipitation occur when 500 mL of a 0.02M solution of AgNO3 is mixed with 500 mL of a 0.001M solution of NaCl? Show your
Oksanka [162]
We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate
7 0
3 years ago
How many moles of S are in 35.4 g of (C3H5)2S?
alexandr402 [8]
<span>0.310 moles First, look up the atomic weights of the elements involved. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight sulfur = 32.065 Molar mass (C3H5)2S = 6 * 12.0107 + 10 * 1.00794 + 32.065 = 114.2086 g/mol Moles (C3H5)2S = 35.4 g / 114.2086 g/mol = 0.309959145 mol Since there's just one sulfur atom per (C3H5)2S molecule, the number of moles of sulfur will match the number of moles of (C3H5)2S which is 0.310 when rounded to 3 significant digits.</span>
6 0
3 years ago
An aqueous solution contains 0.29 M of benzoic acid (HA) and 0.16 M of sodium benzoate (A-). If the pH of this solution was meas
Inga [223]

Answer:

pKa = 4.89.

Explanation:

We can solve this problem by using the <em>Henderson-Hasselbach equation</em>, which states:

pH = pKa + log \frac{[A^-]}{[HA]}

In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

We <u>input the given data</u>:

4.63 = pKa + log \frac{0.16}{0.29}

And <u>solve for pKa</u>:

pKa = 4.89

3 0
3 years ago
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