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saw5 [17]
3 years ago
5

I'll give brainliest

Chemistry
1 answer:
o-na [289]3 years ago
6 0

Answer:

COMPOSITION

Explanation:

Brainliest plz

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I need help on this please
tino4ka555 [31]

Answer:

The answer is c

Explanation:c

3 0
2 years ago
Describe the similarities and differences between herbivores,carnivore,and omnivores​
vredina [299]

Answer:

Herbivores only eat plants . Remember HERBS are PLANTS.

Carnivores only eat meat. Remember CARNE = MEAT in Spanish.

Omnivores eat MEAT & PLANTS.

Explanation:

Differences are what they eat, similarities omnivores eat both.

3 0
2 years ago
How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?
slava [35]

Answer:

A. 0.90 L.

Explanation:

  • NaOH solution will react with H₂SO₄ according to the balanced reaction:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>

<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>

  • For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>

4 0
3 years ago
In the wintertime, you decide to buy a helium balloon at Dollar Tree. It is fully
Gemiola [76]

Answer:

The reason that the balloon shrinks is because when the temperature drops, helium becomes more dense then it already is. So what happens is that its molecules lose energy, slows down and moves closer together to conserve heat and this decreases the volume inside the balloon because the helium molecules are moving closer together, rather than outward toward the shell of the balloon which cause the balloon to swell then shrink. The gas law that causes this to happen is the Charles law.

Explanation:

5 0
3 years ago
Read 2 more answers
A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
3 0
3 years ago
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