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Aleks [24]
3 years ago
9

What are all examples of biotic factors running log, stream water, dead leaves, bacteria, soil

Chemistry
2 answers:
lutik1710 [3]3 years ago
8 0

Im not sure but I think it's rotting log, bacteria, and dead leaves Explanation:

Bumek [7]3 years ago
3 0

dead leaves i think if im not mistaken

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A 20.0 milliliter sample of 0.200 molar K₂CO₃ solution is added to 30.0 milliliters of 0.400 molar Ba(NO₃)₂ solution. Barium car
sveticcg [70]

Answer:

0.32M

Explanation:

<u>Step 1:</u>  Balance the reaction

K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3

We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution

SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3

<u>Step 2: </u>Calculate concentration

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>

<u />

<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3

[Ba2+] = 0.32 M

The concentration of Barium ion in solution is 0.32 M

6 0
3 years ago
PLEASE HELP<br> I really need someone to answer this!!!!
marshall27 [118]

Answer: I think it’s the first one

6 0
3 years ago
Read 2 more answers
Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g)
Marrrta [24]

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

                         3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

                            3(46) kg--------------   2(63) kg  

                                  x     ---------------  7600000 kg

           x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30            

                         2 NO(g) + O2(g)---2 NO2(g)

                       2(30) ------------------2(46)

                        x        ---------------- 8323809.5 kg  

             x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

                      4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

                       4(17) -------------------- 4(30)

                         x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

4 0
3 years ago
What is the pH of a solution with a 3.2 x 10−5 M hydronium ion concentration? (5 points)
stiks02 [169]
B. 4.5

I looked it up
7 0
2 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0
3 years ago
Read 2 more answers
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