All categories

4 years ago

Choose the abiotic factors that characterize the tundra biome.

2 answers:

3 0

The only answer that makes sense is A.

Warm wet forests is a rainforest, arctic foxes are biotic, not abiotic, and little rainfall extreme temperatures is a desert.

olganol [36]4 years ago

3 0

The answer is since abiotic factors are things that are nonliving and tundras are very flat and cold.

You might be interested in

5. Name five plants and animals that lived in ancient Antarctica

Andrew [12]

Answer:

Two hundred million years ago, Antarctica was a lush, temperate rainforest home to crocodile-sized amphibians and rhinoceros-sized dinosaurs.

Explanation:

Animals:

1 ankylosaurs (the armored dinosaurs), mosasaurs and plesiosaurs (both marine reptilian groups).

2 Cry-oloph-os-auru-s was the ap-ex pre-da-tor of its region when it hunted in the lands of Jur-as-sic Period Antarctica.

3 plesiosaurs.

4 mosasaurs.

Plants:

1 palm trees.

2 ferns and Tanglefoot.

3 Evergreen trees

4 Glossopteris

8 0

2 years ago

During the light-independent reaction, carbon dioxide is fixed by adding it to a A. 2-carbon compound B. 3-carbon compound C. 4-

Scrat [10]

During the light independent reaction, carbon dioxide is fixed by adding it to a <span>5-carbon compound</span>

8 0

3 years ago

Read 2 more answers

What is the pOH of a 0.0072 M solution of sodium hydroxide? pOH=?

Rus_ich [418]

POH of a 0.0072 M=-lg(0.0072) = 2.1426675

8 0

3 years ago

Read 2 more answers

In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what

melomori [17]

Answer:

$Y=65.7\%$

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

$n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2$

Thus, we compute the percent yield as:

$Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%$

Best regards.

8 0

3 years ago

What mass of a 0.583 molar solution of iron(III) nitrate is needed to obtain a) 0.0200 moles of iron(III) nitrate, b) 0.0500 mol

geniusboy [140]

Answer:

adfgbhnjm

Explanation:

34rg

8 0

3 years ago