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kipiarov [429]
3 years ago
14

Identify three factors that characterize aquatic systems

Chemistry
1 answer:
givi [52]3 years ago
3 0
While terrestrial biomes are shaped by air temperature and precipitation, aquatic systems are characterized by factors such as water salinity, depth, and whether the water is moving or standing. If that's what you mean?
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glacial striations in the bedrock of south africa indicate that glaciers once moved across the african continent the presence of
finlep [7]
Okay, so the answer is
8 0
2 years ago
The notation for the nuclide 13755Cs gives information about
k0ka [10]

Answer is (3) both mass number and atomic number.

The notation is ₅₅¹³⁷Cs. The Cs represents the chemical symbol of Caesium element. The subscript number at the left hand side of the symbol indicates the atomic number. Hence, atomic number of Cs is 55. The superscript number at the left hand side of the symbol shows the mass number. Hence, the mass number of the Cs is 137.

5 0
3 years ago
2C2H2(g)+5O2(g)=4CO2(g)+2H2O(g)
balandron [24]

The volume of ethyne, C₂H₂ required to produce 12 moles of CO₂ assuming the reaction is at STP is 134.4 L

<h3>Balanced equation</h3>

2C₂H₂(g) + 5O₂(g) --> 4CO₂(g) + 2H₂O(g)

From the balanced equation above,

4 moles of CO₂ were produced by 2 moles of C₂H₂

<h3>How to determine the mole of C₂H₂ needed to produce 12 moles of CO₂</h3>

From the balanced equation above,

4 moles of CO₂ were produced by 2 moles of C₂H₂

Therefore,

12 moles of CO₂ will be produce by = (12 × 2) / 4 = 6 moles of C₂H₂

<h3>How to determine the volume (in L) of C₂H₂ needed at STP</h3>

At standard temperature and pressure (STP),

1 mole of C₂H₂ = 22.4 L

Therefore,

6 moles of C₂H₂ = 6 × 22.4

6 moles of C₂H₂ = 134.4 L

Thus, we can conclude that the volume of C₂H₂ needed for the reaction at STP is 134.4 L

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

3 0
2 years ago
Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equ
Kryger [21]

Answer:

See explanation

Explanation:

The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.

2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.

As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than  2-methylpropane.

3 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
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