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3 years ago

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

Chemistry

2 answers:

castortr0y [4]3 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]3 years ago

6 0

The answer is truer

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sdas [7]

true.

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3 years ago

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A bacteria culture begins with 15 bacteria which double in amount at the end of every hour. Solve for the number of bacteria tha

pshichka [43]

The number of bacteria is given by:
N(t) = N(o) x 2ⁿ
Where N(t) is the number after n hours have passed and N(o) is the original number which is 15.
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3 years ago

What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?

rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

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2 years ago

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Nookie1986 [14]

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3 years ago

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

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3 years ago