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Snezhnost [94]
3 years ago
15

Explain how soil can affect the composition of the solution that move through it

Chemistry
1 answer:
sertanlavr [38]3 years ago
5 0

Answer: Soils are formed through the interaction of five major factors: time, climate, parent material, topography and relief, and organisms.

Soil structure affects plant growth in many, often surprising, ways. The most obvious effects are on root growth, which is strongly inhibited by hard soil, and which in turn influences the ability of the root system to extract adequate water and nutrients from the soil.

Explanation:

https://www.publish.csiro.au/sr/pdf/SR9910717

This is where I gather some info.  

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Which is true about all the different types of matter?
Marat540 [252]
Answer: 1. It is made up of the same basic particles

Matter is made up of atoms, molecules, and ions that cause it to have mass and volume. The different types of matter are made up of the same basic particles but differ on the molecular arrangement and energy the atoms contains that's why they appear differently and have different properties.
8 0
3 years ago
The change in internal energy for the combustion of 1.0 molmol of octane at a pressure of 1.0 atmatm is -5084.1 kJkJ . Part A If
Dovator [93]

Answer:

W = -10.3 kJ

Explanation:

During combustion, the system performs work and releases heat. Therefore, the change in internal energy is negative, and the change in enthalpy, which is equal to heat at constant pressure, is also negative. Work is then calculated by rearranging the equation for the change in internal energy:

w=ΔE−qp=−5084.3 kJ−(−5074.0 kJ)

The release of heat is much greater than the work performed by the system on its surroundings. The potential energy stored in the bonds of octane explains why considerably large amounts of energy can be lost by the system during combustion.

3 0
2 years ago
A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
scoray [572]

Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

  • The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

  • 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
  • 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
5 0
3 years ago
A 100.0mL bubble of hot gases at 225 C and 1.80 atm escapes from an active volcano, what is the new volume of the bubble outside
Inessa05 [86]
<h3>Answer:</h3>

112.08 mL

<h3>Explanation:</h3>

From the question we are given;

  • Initial volume, V1 = 100.0 mL
  • Initial temperature, T1 = 225°C, but K = °C + 273.15

thus, T1 = 498.15 K

  • Initial pressure, P1 = 1.80 atm
  • Final temperature , T2 = -25°C

                                     = 248.15 K

  • Final pressure, P2 = 0.80 atm

We are required to calculate the new volume of the gases;

  • According to the combined gas law equation;

\frac{P1V1}{T1}=\frac{P2V2}{T2}

Rearranging the formula;

V2=\frac{P1V1T2}{T1P2}

Therefore;

V2=\frac{(1.80atm)(100mL)(248.15K)}{(498.15K)(0.80atm)}

V2=112.08mL

Therefore, the new volume of the gas is 112.08 mL

8 0
3 years ago
Rank these species by their ability to act as an oxidizing agent. cu+ mg2+ fe2+
natita [175]

Explanation:

The species or elements which gain electrons and reduces itself are known as oxidizing agent or oxidant.

Ability of an element to act as an oxidizing agent depends on its electrode potential.

The electrode potential of Cu^{+} is 0.52 V.

The electrode potential of Fe^2+ is -0.41 V.

The electrode potential of Mg^{2+} is -2.38 V.

Greater is the value of electrode potential, stronger will be the oxidizing agent.

Therefore, rank of these species by their ability to act as an oxidizing agent are as follows.

       Cu^{+} > Fe^2+ > Mg^{2+}

7 0
3 years ago
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