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Verdich [7]
3 years ago
9

Classify each of the following pure substances as an element or a compound. a. helium gas (He) b. mercury (Hg) in a thermometer

c. sugar (C12H22O11) d. sulfur (S) e. lye (NaOH)
Chemistry
1 answer:
olga nikolaevna [1]3 years ago
4 0

Answer:

A. Element

B. Element

C. Compound

D. Element

E. Compound

Explanation:

A, B, and D are elements while C and E are compounds. The three are elements backside basically they are yet to bond with any other element to form a compound.

Although the mercury is in glass, this doesn’t change the fact that it is an element. Being in glass does not affect the independence it enjoys as an element. This serves to mean that even elements can be useful in their elemental states and this does not affect their initial state. They still remain as elements even if they are in a particular confinement.

The other two I.e lye and sugar are compounds as they contain more than one element which are mixed together in appropriate volume

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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
7 0
3 years ago
A solution labeled "0.105 M NaOH" would contain ______________ moles of NaOH in each liter of solution.
koban [17]
It would contain 0.105M moles of NaOH in each liter of solution
5 0
3 years ago
An atom of element A has 108 protons. What would the elements atomic number be?
marusya05 [52]

108 is the atomic number

3 0
3 years ago
Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.
Shtirlitz [24]

Answer:

(a) pH = -Log (0.1M) = 1

(b) pH = -Log (10^{-13}M) = 13

(c) pH = -Log (3x10^{-3}M) = 2.5

(d) pH = -Log (4.93x10^{-10}M) = 9.3

(e) pH = -Log (5^{-7}M) = 6.3

Explanation:

To calculate de pH of an acid solution the formula is:

pH = -Log ([H^{+}]) = 1

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

K_{w} =[H^{+} ][OH^{-}]=10^{-14}

clearing the [H^{+} ]

[H^{+} ]=\frac{10^{-14}}{[OH^{-}]}

(d) is a weak base so it is necessary to solve the equilibrium first, knowing Ka=3.24x10^{-8}

The reaction is HClO  →  H^{+} + CO^{-} so the equilibrium is

Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}

clearing the <em>x</em>

{x^{2}={1.62x10^{-17}-3.24x10^{-8}x}

x=[H^{+}]=4.93x10^{-10}

6 0
3 years ago
Results published in journals provide the most reliable information about new discoveries because _____.
sesenic [268]

The correct answer is D, they are reviewed by experts.

<em>(Please mark this answer as Brainliest and leave a Thanks if I helped you!)</em>

3 0
4 years ago
Read 2 more answers
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