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aleksklad [387]
3 years ago
7

What data would be most helpful to support the claim that the population of a certain plant is dependent on the rainfall in its

habitat
1. Rainfall amounts and number of these plants in a certain habitat over several years

2. The heights and number of leaves produced by these plants under different moisture conditions

3. The average amount of water absorbed by the roots and given off by the leaves of these plants

4. The amount of water retained in the soil in the area where the plants live
Chemistry
1 answer:
LekaFEV [45]3 years ago
5 0

Answer: number 1 rainfall amounts

Explanation: i hope this heps you bye

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What is energy
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it is s because enegy allows us to walk

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Which best defines partial pressure in a mixture of gases?
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Pressure that is exerted by one gas as if it occupied a container by itself.

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Calculate the molar mass of each of the following:
Allushta [10]

Explanation:

Molar mass

The mass present in one mole of a specific species .

The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .

(a) S₈

Molar mass of of the atoms are -

sulfur, S = 32 g/mol.

Molar mass of  S₈ = 8 * 32 g/mol.  = 256 g/mol.

(b) C₂H₁₂

Molar mass of of the atoms are -

Hydrogen , H = 1 g/mol

Carbon , C = 12 g/mol

Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol

(c) Sc₂(SO₄)₃

Molar mass of of the atoms are -

sulfur, S = 32 g/mol.

oxygen , O = 16 g/mol.

scandium , Sc = 45 g/mol.

Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol  

(d) CH₃COCH₃ (acetone)

Molar mass of of the atoms are -

Carbon , C = 12 g/mol

oxygen , O = 16 g/mol.

Hydrogen , H = 1 g/mol

Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol

(e) C₆H₁₂O₆ (glucose)

Molar mass of of the atoms are -

Carbon , C = 12 g/mol

oxygen , O = 16 g/mol.

Hydrogen , H = 1 g/mol

Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.

6 0
3 years ago
under certain water​ conditions, the free chlorine​ (hypochlorous acid,​ hocl) in a swimming pool decomposes according to the la
ivolga24 [154]

Answer:

1.7 ppm

Explanation:

Original amount N' = 2.6 ppm

time to testing t = 24 hr

final amount N = 2.1 ppm

Using exponential inhibited decay, we have

N = N'e^(-kt)

Where

N is the new reading

N' is the original reading

t is the decay time

k is the decay constant

Substituting, we have

2.1 = 2.6 x e^(-k x 24)

2.1 = 2.6 x e^(-24k)

0.808 = e^(-24k)

We take the natural log of both sides of the equation

Ln 0.808 = Ln (e^(-24k))

-0.213 = - 24k

K = 0.213/24 = 0.00886

After 48 hrs, the reading of free chlorine will be

N = 2.6 x e^(-0.00886 x 48)

N = 2.6 x e^(-0.425)

N = 2.6 x 0.654

N = 1.7 ppm

5 0
3 years ago
3
Murrr4er [49]

Answer:

Look at the properties of Oxygen and Silicon - the two most abundant elements in the Earth's crust - by clicking on their symbols on the Periodic Table.

Explanation:

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