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Cerrena [4.2K]
2 years ago
8

Describe how organisms in the tropical rainforest both depend on and compete for biotic and abiotic factors.

Chemistry
1 answer:
Karolina [17]2 years ago
6 0

Answer: An organism's niche includes food, shelter, its predators, the temperature, the amount of moisture the organism needs to survive, etc. When two or more individuals or populations try to use the same limited resources such as food, water, shelter, space, or sunlight, it is called competition.

Explanation: Hope that helps?

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PLEASE HELP ASAP
Leni [432]

Answer:

the answer is B

Explanation:

the answer is B

4 0
3 years ago
Read 2 more answers
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
How many grams are there in 457.25 pounds?
Ne4ueva [31]
There are 207405.111 grams in that many pounds.
7 0
4 years ago
Read 2 more answers
What are the answers to these?
antiseptic1488 [7]

Answer: 1. AgF + CaCl2 = AgCl + CaF2

2. C2H4 +O2 = CO2 +H2O

3. K2S = K+S

4. O2 + Mg = MgO

5. Mg + AlBr3 = MgBr2 + Al

6.C2H6O + O2= CO2 + H2O

7.Li2SO4 + MgCl2= Li2SO4 + MgCl2

8.HCl + Zn= H2 + ZnCl2

Explanation:

Balance the equation

Write down your given equation.

Write down the number of atoms per each element that you have on each side of the equation.

Always leave hydrogen and oxygen for last.

If you have more than one element left to balance:

Add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation.

Balance the hydrogen atoms next.

Balance the oxygen atoms.

4 0
4 years ago
What mass of CO was used up in the reaction with an excess of oxygen gas if 24.7g of carbon dioxide is formed? 2 CO + O2 &gt; 2
Zolol [24]
Balance Chemical Equation,
                                      2 CO  +  O₂   →   2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when  =  56 g (2 mole)of CO was reacted
So, 
24.7 g of CO₂ will be produced by reacting =  X g of CO

Solving for X,
                                    X  =  (56 g × 24.7 g) ÷ 88 g

                                    X  =  2.26 g ÷ 88 g

                                    X  =  0.0257 g of CO

Result:           
            0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
7 0
3 years ago
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