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nekit [7.7K]
3 years ago
8

which feature of the ocean floor includes its deeper parts ? A.Trench B.estuary C.mid-ocean D. contnetal slop

Chemistry
1 answer:
morpeh [17]3 years ago
7 0

Answer:

A

has to 20 characters lol

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URGENTTTTT HELPPPP PLZZZ LAST TRYYY
algol13

Left Panel

A is an acid. Not the answer.

B is correct. That would be a base. But it is not an Arrhenius base. Keep reading.

C that is exactly what an Arrhenius base is.

D. No an acid of some sort would accept OH ions.

Right Panel

D is concentrated and it is also a weak base. Good cleaning fluid. Smells awful but it works.

8 0
2 years ago
What is needed to change the state of matter
almond37 [142]
Heat is needed to change solid to liquid even to gas!
6 0
3 years ago
Read 2 more answers
Rn-222 has a half-life of 3.82 days. If 25.0 g of Radon-22 was originally present, approximately how many grams would be left af
horsena [70]

Answer:

Approximately 0.39 g or 0.4 g if you're rounding up

Explanation:

15/3.82 = 3.92

Let's round that up to 4

That means 15 days is around 4 half lives

4 half lives means 1/16 of the original mass will be left

25/16 = 0.390625

6 0
3 years ago
Iron iii chloride ammonium hydroxide balanced equation, complete ionic equation, net ionic equation
Vladimir [108]
Iron III Chloride has a chemical formula of FeCl₃, while ammonium hydroxide has a chemical formula of NH₄OH. 

The <em>balanced equation</em> would be:

FeCl₃ (aq) + 3 NH₄OH (aq) → Fe(OH)₃ (s) + 3 NH₄Cl (aq)
The precipitate is Fe(OH)₃ or iron iii hydroxide.

To find the <em>complete ionic equation</em>, dissociate the compounds in aqueous phases into their ionic forms:

Fe³⁺ + Cl⁻ + NH₄⁺ + 3 OH⁻ -->  Fe(OH)₃(s) + NH₄⁺ + Cl⁻ 

To find the <em>net ionic equation</em>, cancel out like ions that appear both in the reactant and product side:

Fe³⁺ +  3 OH⁻ -->  Fe(OH)₃


4 0
3 years ago
If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?
bija089 [108]
<h3>Molar mass:-</h3>

\\ \sf\longmapsto HC_2H_3O_2

\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)

\\ \sf\longmapsto 1u+24u+3u+48u

\\ \sf\longmapsto 28u+48u

\\ \sf\longmapsto 76u

\\ \sf\longmapsto 76g/mol

  • No of moles=0.2mol
  • Given mass=?

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}

\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}

\\ \sf\longmapsto Given\:Mass=0.2\times 76

\\ \sf\longmapsto Given\:Mass=1.52g

7 0
3 years ago
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