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Neko [114]
3 years ago
12

Which is not an example of a compound? Ingnore the top PLEASE HELPP

Chemistry
2 answers:
Mademuasel [1]3 years ago
7 0

Answer:

The 3rd answer or NCI is the correct answer

Explanation:

Hope this helps:)

ZanzabumX [31]3 years ago
3 0

Answer:

the 1st one

Explanation:

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The PH of a solution of Hcl is 2.find out the amount of acid present in a litre of the solution ​
Scrat [10]

Answer:

The solution is 10^-2 or 0.01M in HCl.

Explanation:

meaning of pH is "power of hydrogen".

what is the molar concentration of a HCl solution with pH=2?

Let say pH=2

[H+]=10^-2M

HCL is a strong acid that dissociates completely:

[H+]=[HCL]

Therefore solution is 10^-2 or 0.01M in HCL.

5 0
3 years ago
PLS HELP Students in a chemistry class added 5 g of Zinc (Zn) to 50 g of hydrochloric acid (HCl). A chemical reaction occurred t
Vitek1552 [10]

Answer: In simplest case mass of reactants is same as mass of products.

Without thinking this question deeper, mass of ZnCl2 would be 49, but..

Explanation: Reaction should be  Zn + 2 HCl ⇒ ZnCl2 + H2

Amount of zinc is  5 g / 65,38 g/mol = 0,076476 mol and amount

of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.

Althought HCl is needed 0.152 moles, zinc is an limiting reactant.

So it is possible to produce only 0.076476 mol Hydrogen and its mass

is 0.154 g.  Mass of ZnCl2 would be 0.076476 mol · (65.38 + 2·35.45) =

 10.42 g

4 0
3 years ago
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Harman [31]
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y

3y = 1.2
y = 0,4M

Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4

C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄

mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol

164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
5 0
4 years ago
Permanently frozen soil is called what?
Kazeer [188]
It is called permafrost :)
8 0
3 years ago
Read 2 more answers
Calculate the solubility of silver oxalate, ag2c2o4, in pure water. ksp = 1.0 × 10-11
aksik [14]
Answer is: the solubility of silver oxalate is <span>a. 1.4 × 10-4 m.
</span>

<span>Chemical reaction (dissociation) of silver oxalate in water: 
Ag</span>₂C₂O₄(s) → 2Ag⁺(aq) + C₂O₄²⁻<span>(aq).
Ksp(Ag</span>₂C₂O₄) = [Ag⁺]²·[C₂O₄²⁻<span>].
[C</span>₂O₄²⁻] = x; solubility of oxalate ion.

[Ag⁺] = 2[C₂O₄²⁻<span>] = 2x
1.0·10</span>⁻¹¹ = (2x)² · x = 4x³.

x = ∛1.0·10⁻¹¹ ÷ 4.

x = 1.4·10⁻⁴ M.

6 0
3 years ago
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