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kirill115 [55]
3 years ago
12

A certain plant requires moisture oxygen carbon dioxide light and minerals in order to survive.This statement describes a living

organism that depends on ______. A.carnivore or herbivore relationships B. biotic factors C.abiotic factors or D.living factors
Chemistry
1 answer:
TEA [102]3 years ago
3 0
The Answer is C. Abiotic Factors
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Write a conversion factor that converts between moles of calcium ions in calcium phosphide, Ca3P2, and moles of Ca3P2.
cricket20 [7]

The conversion factor between moles of calcium ion and moles of Ca_3P_2 in  Ca_3P_2 would be 3 to 1.

<h3>Mole fraction</h3>

Ca_3P_2 if formed from 3 carbon atoms and 3 phosphate atoms according to the equation:

3Ca + 2P --- > Ca_3P_2

Thus, there are 3 moles of calcium in every 1 mole of  Ca_3P_2.

The conversion factor between moles of calcium and moles of  Ca_3P_2 will. therefore, be 3 to 1 or simply 3:1.

More on mole fractions can be found here: brainly.com/question/8076655

#SPJ1

6 0
3 years ago
What environmental factors might make a strong odor an advantage for the
Molodets [167]

Answer:

Rafflesia smell like roadkill others for only a few hours and while a few smaller species barely smell at all. And In short  a parasitic life is heavily specialized biological requirements.

Explanation:

6 0
3 years ago
Help pls<br>it would be really helpful
Basile [38]
18.is D
19.is A
hope it's correct
3 0
3 years ago
How many moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm
Citrus2011 [14]

<u>We are given:</u><u>_______________________________________________</u>

Volume of Gas (V) = 2.5L

Pressure (P) = 1.2 atm

Temperature (T) = 25°C  OR 25+273 = 298 K

Universal Gravitational Constant (R) = 0.0821

<u>Solving for number of moles:</u><u>___________________________________</u>

From the Ideal Gas Equation,

PV = nRT

(1.2)(2.5) = n(0.0821)(298)           [plugging the given values]

n = [(1.2)(2.5)] / [0.0821*298]

n = 300 / [298*8.21]

n = 0.12 moles

Hence, there are 0.12 moles of Oxygen in 2.5L of 1.2 atm gas when the temperature is 25°C

7 0
3 years ago
What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
ioda

N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

Masses transformed into moles:

Formula used

Given mass/ Molar mass

14.01 g contains 1 mol of N

So 63.61 g of N contains moles is equals to

(1 mol N / 14.01 g N) 63.61 g N = 4.540 mol N

Similarly

16 g of O contains 1 mole of O

36.69 g of O contains moles is equals to

(1 mol O / 16.00 g O) 36.69 g O = 2.293 mol O

Divide by the smallest to normalize:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

Therefore, there are roughly twice as many N as O atoms. N2O is the empirical formula as a result.

Ratio is basically 2:1

Hence, N₂O is the empirical formula of an oxide of nitrogen

Learn more about Empirical Formula here brainly.com/question/27873410

#SPJ4

7 0
2 years ago
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