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kvv77 [185]
2 years ago
11

Which choices is a biotic factor of an ecosystem? • sun • ants • air • soil

Chemistry
1 answer:
mojhsa [17]2 years ago
8 0

Answer:

Aboitic factor would be soil

Biotic factor would be ants.

Explanation:

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2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme
mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O

Next, we assign the oxidation numbers as follows:

2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA requ
egoroff_w [7]

Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

<u>Step 1:</u> Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

<u>Step 2:</u> Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+  in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed  in the back titration with Ca2+:

<u>Step 3: </u>Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

<u>Step 4:</u> Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

<u>Step 5:</u> Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction  with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

<u>Step 6:</u> Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

<u>Step 7: </u>Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

4 0
3 years ago
Chemical Quantities "Molar Mass - 2 step" - Wksh #3 Directions: Use dimensional analysis to perform the following calculations.
fomenos

There are about 1.27  × 10^23 molecules present in 13.5 g of SO2.

<h3>What are molecules?</h3>

A molecule is the simplest part of a compound that have independent existence.

Number of moles of sulfur dioxide = 13.5 g/64 g/mol = 0.211 moles

If 1 mole of SO2 contains 6.02 × 10^23 molecules

0.211 moles of SO2 contains 0.211 moles × 6.02 × 10^23 molecules/1 mole

= 1.27  × 10^23 molecules

32 g of sulfur contains 6.02 × 10^23 atoms

x g of sulfur contains  2.23 x 10^23

x = 32 × 2.23 x 10^23 /6.02 × 10^23

x = 11.85 g

6.02 × 10^23 formula units has a mass of AgF = 127 g

x formula units has a mass of  = 42.15 g

x = 6.02 × 10^23 × 42.15/127

x = 2 × 10^23 formula units

6.02 × 10^23 formula units of Fe2O3 has a mass of 160 g

8.83 x 10^23 formula units of Fe2O3 has a mass of 8.83 x 10^23 x 160 g/6.02 × 10^23

x = 235 g

Learn more about formula unit: brainly.com/question/19293051

6 0
2 years ago
As a chlorine atom becomes a negative ion, the atom blanks an electron
Andrew [12]

Answer: increases by

As a chlorine atom becomes a negative ion, the atom "increases by" an electron

Explanation:

Chlorine atom has an atomic number of 17, and an electronic configuration of 1s2, 2s2 2p6, 3s2 3p5 showing 7 valence electrons in its outermost shell.

Hence, it receives a single electron to achieve a stable octet structure with electronic configuration of

1s2, 2s2 2p6, 3s2 3p6. Therefore, Cl- is a univalent negative ion with 8 valence electrons in its outermost shell. The increase is shown below

Cl + e- --> Cl-

Thus, as a chlorine atom becomes a negative ion, the atom "increases by" an electron

7 0
3 years ago
What is balance equation
satela [25.4K]
Search it up on google
7 0
4 years ago
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