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steposvetlana [31]

4 years ago

11

Earth’s cycles provide organisms with continued access to usable elements necessary for sustaining life. (True or False)

2 answers:

castortr0y [4]4 years ago

8 0

Answer:

True

Explanation:

There are several physical, chemical , biological and chemical –biological cycles in the earth’s atmosphere that keep on rotating the minerals with in the different segments of earth. There are several such cycles –

a) Nitrogen cycle

b) Carbon cycle

c) Phosphorous cycle

d) Food cycle

e) Energy cycle ( trophic levels in a food web)

The primary constituents of all these cycle keep on changing their form from species to species so that these nutrient get available to all the living beings.

tensa zangetsu [6.8K]4 years ago

6 0

The answer is truer

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How does the structure of a carbon atom enable it to form large molecules

zlopas [31]

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3 years ago

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Calculate the amount of heat needed to boil 64.7 g of benzene (C6H6), beginning from a temperature of 41.9 C . Round your answer

JulsSmile [24]

Answer: The amount of heat needed is = 4.3kJ

Explanation:

Amount of heat H = M × C × ΔT

M= mass of benzene = 64.7g

C= specific heat capacity = 1.74J/gK

ΔT = T2-T1

Where T1 is initai temperature = 41.9C

T2 is the final temperature( boiling point of benzene) = 80.1C

H= 64.7×1.74×80.7

H= 4300J

H=4.3kJ

Therefore, the amount of heat needed is 4.3kJ

8 0

3 years ago

Steel is the most commonly used metallic material. However, its annual loss due to corrosion is huge. Preventing metal corrosion

Fiesta28 [93]

Answer:

d

Explanation:

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5 0

3 years ago

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Consider the dissolution of AB(s): AB(s)⇌A+(aq)+B−(aq) Le Châtelier's principle tells us that an increase in either [A+] or [B−]

Arlecino [84]

Answer:

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. $3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$ .

Explanation:

A

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

S 2S

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$K_{sp}=S\times (2S)^2=4S^3$

$8.45\times 10^{-12}=4S^3$

Solving for S:

$S=0.000128 M$

0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of $M(NO_3)_2$ = 0.202 M

Solubility product of generic metal hydroxide = $K_{sp}=8.45\times 10^{-12}$

$M(OH)_2\rightleftharpoons M^{2+}+2OH^-$

S 2S

So, $[M^{2+}]=0.202 M+S$

The expression of a solubility product is given by :

$K_{sp}=[M^{2+}][OH^-]^2$

$8.45\times 10^{-12}=(0.202 M+S)(2S)^2$

Solving for S:

$S=3.23\times 10^{-6} M$

$3.23\times 10^{-6} M$ is the solubility of $M(OH)_2$ in a 0.202 M solution of $M(NO_3)_2$ .

8 0

4 years ago

What is the approximate energy of a photon having a frequency of 4 x 10^7 Hz (h=6.6 x 10^-34]•s)

jok3333 [9.3K]

Answer:

E = 26.4 ×10⁻²⁷ j

Explanation:

Given data:

Energy of energy =?

Frequency = 4×10⁷ Hz

Solution:

Formula:

E = hf

E = hf

E = 6.6 ×10⁻³⁴ m²Kg/s . 4×10⁷ Hz

Hz = s⁻¹

j = m²Kg/s ²

E = 26.4 ×10⁻²⁷ j.

5 0

4 years ago