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NikAS [45]
2 years ago
11

How does an ecosystem benefit when leaves fall from a tree and decompose on the soil?

Chemistry
2 answers:
nexus9112 [7]2 years ago
5 0

Answer:

A. Nutrients are released into the soil

miskamm [114]2 years ago
4 0
The first answer is w
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levacccp [35]

Answer:

A pure substance has a definite chemical composition

4 0
3 years ago
CHM 17 Practice Final Exam
serious [3.7K]

Answer:

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Explanation:

7 0
3 years ago
What is the mass in grams of 1.5*10^-3 mol of sodium
Nutka1998 [239]
The Formula to find the number of moles is:
Number of Moles = Mass/ Molar Mass

Which means that Mass = Number of Moles * Molar Mass

We know that Number of Moles of Sodium Na = 1.5*10^_{-3} moles 
If you check the Periodic Table, Molar Mass of Sodium Na = 23g/mol

So, Mass = 1.5*10^_{-3} * 23 = 3.45 * 10^{-2} 

So, the mass in grams of  1.5*10^_{-3} mol of Sodium is 3.45 * 10^{-2}

Hope this Helps :)
4 0
3 years ago
Read 2 more answers
Calculate the value of ΔG°, ΔH°, & ΔS° for the oxidation of solid elemental sulfur to gaseous sulfur trioxide at at 25°C.​2S
Liula [17]

Answer:

ΔG° = -750324.96 J/mol = -750.324 Kj/mol

ΔH° = - 790.4 kJ/mol = -790400 j/mol

ΔS° =  = - 134.48 J K-1mol-1

Explanation:

(S, rhombic) + 3O2( g) → 2SO3 (g)

Temperature = 25°C + 273 = 298K (Converting to kelvin temperature.)

The formulae relating all four paramenters (ΔG°, T, ΔH°, & ΔS°) is given as;

ΔG° = ΔH° - TΔS°

All H and S values used are measurements at 25°C

Calculating ΔH

You started with 1 mole of Sulphur and 3 moles of oxygen.

Total starting enthalpy = 0+ 3(0) = 0 kJ/mol

You ended up with 2 moles of SO3.

Total enthalpy at the end = 2(-395.2) = - 790.4 kJ/mol

enthalpy change = what you end up with - what you started with.

ΔH = enthalpy change =  - 790.4 kJ/mol - 0 kJ/mol = - 790.4 kJ/mol = -790400 j/mol

Calculating ΔS

You started with 1 mole of Sulphur and 3 moles of oxygen.

Total starting entropy =31.88 + 3(205) = 646.88 J K-1mol-1

You ended up with 2 moles of SO3.

Total entropy  at the end = 2(256.2) = 512.4 J K-1mol-1

entropy change = what you end up with - what you started with.

ΔS = entropy change =  512.4 J K-1mol-1 - 646.88 J K-1mol-1

= - 134.48 J K-1mol-1

Inputing the values into the formular, we have;

ΔG° = ΔH° - TΔS°

ΔG° = - 790400 - 298 (- 134.48)

ΔG° =  - 790400 + 40075.04

ΔG° = -750324.96 J/mol = -750.324 Kj/mol

4 0
2 years ago
Mass is 80g. Volume is 40ml. What is the density
IRISSAK [1]
80 / 40 = 2 Therefore the density is 2g/mL
6 0
2 years ago
Read 2 more answers
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