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Finger [1]
3 years ago
12

ASAP HELP

Chemistry
2 answers:
jeyben [28]3 years ago
7 0

Answer:

the answer is 2.75 but since that isn't an option I'd go with 1 because I just took a test with this question and it was the right answer.. hopefully that helps

BabaBlast [244]3 years ago
6 0

Answer:

2.75

Explanation:

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Sodium chloride is an ionic compound. If you drop a large crystal of sodium chloride on a hard floor. Choices are change to gas,
natita [175]
Shatter into tiny pieces

The crystal may break due to the fall.
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3 years ago
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
zubka84 [21]

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

6 0
3 years ago
What type of reaction is Zn + CuCl2 → Cu + ZnCl2?​
Shalnov [3]

Answer:

Its single - displacement.

Explanation:

4 0
4 years ago
When hydrogen peroxide (H2O2) is added to potassium iodide (KI) solution, the hydrogen peroxide decomposes into water (H2O) and
riadik2000 [5.3K]

Answer:

D. Catalyst

Explanation:

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5 0
3 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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