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3 years ago

Why doesn’t a stable ecosystem have more tertiary consumers than primary consumers?

1 answer:

5 0

Tertiary consumers are usually big predators which
1) have a tendency to claim large territories for themselves
2) need more energy and thus eat more than primary consumers

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Beth heats a pot of water to cook pasta.<br><br><br> What may happen as a result of adding heat?

n200080 [17]

The water will started boiling and the pasta is going to start like melting so it can get a soft and smooth texture so you can eat it.

6 0

3 years ago

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How do her cells use these molecules to release energy for her body to run?

Anika [276]

In cells use oxygen to release energy stored in sugars such as glucose. In fact, most of the energy used by the cells in your body is provided by cellular respiration. Just as photosynthesis occurs in organelles called chloroplasts, cellular respiration takes place in organelles called mitochondria.

6 0

3 years ago

What is the molarity of a nitric acid solution if 43.13 mL 0.1000 M KOH solution is needed to neutralize 30.00 mL of the acid so

Leno4ka [110]

Answer:

0.144M

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

HNO3 + KOH —> KNO3 + H20

From the equation,

nA = 1

nB = 1

From the question given, we obtained the following:

Ma =?

Va = 30.00mL

Mb = 0.1000M

Vb = 43.13 mL

MaVa / MbVb = nA/nB

Ma x 30 / 0.1 x 43.13 = 1

Cross multiply to express in linear form

Ma x 30 = 0.1 x 43.13

Divide both side by 30

Ma = (0.1 x 43.13) /30 = 0.144M

The molarity of the nitric acid is 0.144M

8 0

3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}