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Marina CMI [18]
3 years ago
13

Pine trees slightly prefer slightly acidic soil. What happens if the pH changes to above 7?​

Chemistry
2 answers:
egoroff_w [7]3 years ago
7 0

Answer:

The trees' growth might be affected

Explanation:

A pH above 7 indicates that the soil is no more acidic and that it had turned basic or alkaline in nature. This may affect the functioning of enzymes as this value is very far from optimum pH value. Thus growth is affected.

WARRIOR [948]3 years ago
7 0

Answer:

It means the soil is basic.

Explanation:

The pH scale lies between 0 and 14. The following can be concluded depending on the pH value.

1. If the pH is between 0 and 6, it means it is acidic.

2. If the pH is 7, it means it is neutral.

3. If the pH is between 8 and 14, it means it is basic/alkaline

From the question, we were told that the pH of the soil is above 7. Therefore, the soil is basic.

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The choices can be found elsewhere and as follows:

A)The reaction requires the collision of three particles with the correct energy and orientation. 
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What is the volume at STP of 3.44 x 1023 molecules of CO2
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Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains 6.022 \times 10^{23} molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

P\cdot V = n\cdot R_{u}\cdot T (1)

P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}} (1b)

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P - Pressure, measured in pascals.

V - Volume, measured in liters.

r - Amount of molecules, no unit.

N_{A} - Avogadro's number, no unit.

R_{u} - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 100000\,Pa, r = 3.44\times 10^{23}, N_{A} = 6.022\times 10^{23}, T = 273.15\,K and R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}, then the volume of carbon dioxide at STP is:

V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}

V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}

V = 12.972\,L

Therefore, the correct answer is C.

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