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notsponge [240]
4 years ago
5

What are two roles living things play in soil formation?

Chemistry
1 answer:
Studentka2010 [4]4 years ago
8 0
Soil Formation<span>:- </span>Living<span> Organisms. Plants, animals, and micro-organisms (fungi and bacteria) all affect </span>soil formation<span> by producing or contributing to humus production. The amount of humus in a </span>soil<span> is a result of how much plant material has been incorporated into it.</span>
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How many protons and electrons do the following ion have?
GalinKa [24]

a) Number of protons = number of electron in the atom = atomic number

So it has 7 protons answer

N-3 has gained 3 electron so number of electrons + 7+3

= 10 answer

8 0
4 years ago
In the combustion of ethane, shown above, how many moles of carbon dioxide can be produced from 1.00 mole of ethane? A. 0.500 mo
Daniel [21]
The balanced chemical equation for the <span>combustion of ethane is 
2C</span>₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

The stoichiometric ratio between C₂H₆(g) and CO₂(g)  is 1 : 2

Hence,
    moles of 
CO₂(g) produced = moles of reacted C₂H₆(g) x 2
                                                 
= 1.00 mol x 2
                                                  = 2.00 mol

Hence, the correct answer is "C".
6 0
3 years ago
What the significance of “Er” in the diagram
Vlada [557]

Answer:

D

Explanation:

3 0
3 years ago
Read 2 more answers
Calculate the number of hydrogen atoms in a sample of hydrazine . Be sure your answer has a unit symbol if necessary, and round
gladu [14]

Answer:

atoms \ H= 9.767x10^{24}atoms

Explanation:

Hello!

In this case, considering that the mass of hydrazine is missing, we can assume it is 130.0 g (a problem found on ethernet). In such a way, since we need a mass-mole-atoms relationship by which we can compute moles of hydrazine given its molar mass (32.06 g/mol), then the moles of hydrogen considering one mole of hydrazine has four moles of hydrogen and one mole of hydrogen has 6.022x10²³ atoms (Avogadro's number); therefore, we proceed as shown below:

atoms \ H=130.0gN_2H_4*\frac{1molN_2H_4}{32.06gN_2H_4} *\frac{4molH}{1molN_2H_4} *\frac{6.022x10^{23}atoms}{1molH}\\\\atoms \ H= 9.767x10^{24}atoms

Notice 130.0 g has four significant figures, therefore the result is displayed with four as well.

Best regards!

7 0
3 years ago
In an addition reaction to an alkene the pi bond plays the role of
stealth61 [152]
Nucleophile electrophile
7 0
3 years ago
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