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love history [14]
3 years ago
8

The alternative pathways of photosynthesis using the c4 or cam systems are said to be compromises. why?

Chemistry
2 answers:
frosja888 [35]3 years ago
7 0
 It is because C4 compromises on water loss and CAM compromises on photorespiration. and Both minimize photorespiration but expend more ATP during carbon fixation. 
pentagon [3]3 years ago
3 0

Both minimize photorespiration but expend more ATP during carbon fixation is the correct answer on MasteringBio.

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What is the difference between material resources and energy resources?
Luba_88 [7]
Mineral resources are solid, crystalline substances made inside the Earth. These include granite, marble, limestone and precious stones which are used for jewellery. Minerals are used to make all sorts of different things which we use every day. Energy resources are things we can use to turn into electrical power.
7 0
3 years ago
SOLVE The density of a gas is 0.68 g/mL. What amount of space would 23.8 g of this gas occupy?
Contact [7]

Answer:

<h2>The answer is 35 mL</h2>

Explanation:

Density of a substance can be found by using the formula

Density(\rho) =  \frac{mass}{volume}

From the question we are finding the amount of space the gas will occupy that's the volume of the gas

Making volume the subject we have

volume =  \frac{mass}{Density}

From the question

mass = 23.8 g/mL

Density = 0.68 g/mL

Substitute the values into the above formula and solve

That's

volume =  \frac{23.8}{0.68}

We have the final answer as

<h3>35 mL</h3>

Hope this helps you

6 0
3 years ago
A 1.59 mol sample of Kr has a volume of 641 mL. How many moles of Kr are in a 4.41 L sample at the same temperature and pressure
Marina86 [1]

Answer:

The correct answer is 10.939 mol ≅ 10.94 mol

Explanation:

According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).

For the initial gas (1), we have:

n₁= 1.59 mol

V₁= 641 mL= 0.641 L

For the final gas (2), we have:

V₂: 4.41 L

The relation between 1 and 2 is given by:

n₁/V₁ = n₂/V₂

We calculate n₂ as follows:

n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol

5 0
2 years ago
What is the average yearly rate of change of carbon-14 during the first 5000 years?
erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

A = A_0 \times e^{-\lambda \times t}

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

T_{1/2} = The half life

\lambda = 0.693/T_{1/2}

t = The time elapsed = 5000 years

λ = 0.693/T_{1/2} = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years  is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0
3 years ago
Empirical formula for S3O9
Vsevolod [243]

Answer:

* The empirical formula of a compound shows the ratio of elements present in a compound

* The molecular formula of a compound shows how many atoms of each element are present in a molecule of a compound.

Example: the compound butene has a molecular formula of C4H8. The empirical formula

of butene is CH2 because there is a 1:2 ratio of carbon atoms to hydrogen atoms.

7 0
3 years ago
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