True of False: the only thing plants do in an ecosystem is provide food.

Chemistry

1 answer:

jekas [21]3 years ago

6 0

Explanation:

False, they also provide oxygen

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Question

Romashka-Z-Leto [24]

2. Its showing that Customers who are in the Value X Rewards Program would also receive free popcorn or drinks each time they would shop at the store.

Hope this helped.

7 0

3 years ago

Read 2 more answers

Convert 533 cm/s to units of meters per minute. Show the unit analysis by dragging the conversion factors into the unit‑factor s

lukranit [14]

Answer:

319.8 m/min

Explanation:

533 cm/s

We can convert 533 cm/s to m/min by doing the following:

First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:

Recall:

100 cm/s = 1 m/s

Therefore,

533 cm/s = 533 cm/s /100 cm/s × 1 m/s

533 cm/s = 5.33 m/s

Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:

1 m/s = 60 m/min

Therefore,

5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min

5.33 m/s = 319.8 m/min

Therefore, 533 cm/s is equivalent to 319.8 m/min

8 0

4 years ago

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.