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Leokris [45]
3 years ago
7

Many grasses and few trees characterize ____________ (tundras/temperate grasslands).

Chemistry
2 answers:
Oksanka [162]3 years ago
8 0

Answer:

Temperate Grasslands

Explanation:

Tundras have nothing to do with grass. Tundras are more like ice

kozerog [31]3 years ago
3 0
Temperate grasslands
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Convert 67 inches into a feet
fiasKO [112]

Answer:

Explanation:every feet has 12in so divde, 67/12  so its 5ft 7/12

4 0
3 years ago
10.0 mL of 3.0 M sulfuric acid has been added to 50.0 mL of water.
Cloud [144]

Answer:

the new concentration is 0.60M

Explanation:

The computation of the new concentration is shown below;

We know that

M1V1=M2V2

(3.0M) (10.0 mL) = M2 (50.0mL)

30 = M2 (50.0mL)

So, M2 = 0.60 M

Hence, the new concentration is 0.60M

The same is considered and relevant

3 0
2 years ago
Which organisms in soil helps relieve element so they can be recycled
Rudiy27

Answer:

Bacteria are also involved in many processes that are indispensable for our life on Earth. Many of these processes have to do with the recycling (reuse) of chemical elements that have been here since the formation of the planet

Explanation:

I hope it will help you

3 0
3 years ago
Which of the following reactions does the coenzyme biotin catalyze?
Papessa [141]

Answer:

c. decarboxylation of an a-keto acid.

Explanation:

Decarboxylation refers to the removal of the carboxyl group from a carboxylic acid and thus releasing carbon dioxide. Decarboxylases are enzymes that speed up the removal of the carboxyl group from acids. These reactants could be amino acids, alpha-keto acids, and beta-keto acids. Biotin is known to catalyze the decarboxylation of malonyl CoA to acetyl CoA during fatty acid synthesis.

Malonyl CoA is converted to acetyl CoA after decarboxylation assisted by biotin also known as Vitamin H. Alpha keto acids are involved in fatty acids synthesis and Malonyl CoA is an alpha-keto acid because the keto group is located in the first carbon near the carboxylic acid group. Keto acids have both a carboxyl group and a ketone group.

5 0
3 years ago
If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

1810.04571428[mL]=V_2

Adjusting for significant figures, this gives V_2=1810[mL]

4 0
1 year ago
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