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umka21 [38]
4 years ago
13

Organic matter with a _______ ratio results in a low net release of nutrients during decomposition. This is because microbial gr

owth is more limited by _______ than it is by _______. low C:N; energy; N supply high C:N; N supply; energy high C:H2O; water supply; energy low C:H2O; energy; water supply low N:H2O; N supply; water supply
Chemistry
1 answer:
Serhud [2]4 years ago
6 0

Answer:

high C:N; N supply; energy

Explanation:

Nitrogen supply is required for microbial growth to synthesize nutrients such as amino acids and proteins. For a high C:N ratio, the amount of nitrogen supply is considerably small compared with the amount of carbon. As a result, a low amount of nutrients is released during decomposition.

In the given question, the organic matter with a __high C:N___ ratio results in a low net release of nutrients during decomposition. This is because microbial growth is more limited by __N supply____ than it is by __energy_____.

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Which of the following is not considered a base unit according to the SI units of measurement
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A sample of oxygen gas at a pressure of 1.19 atm and a temperature of 24.4 °C, occupies a volume of 18.7 liters. If the gas is a
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Answer:

\boxed {\boxed {\sf 0.757 \ atm}}

Explanation:

We are asked to find the pressure of a gas given a change in volume. Since the temperature remains constant, we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume is inversely proportional to the pressure. The formula for this law is:

P_1V_1= P_2V_2

Initially, the oxygen gas occupies a volume of 18.7 liters at a pressure of 1.19 atmospheres.

1.19 \ atm * 18.7 \ L = P_2V_2

The gas expands to a volume of 29.4 liters, but the pressure is unknown.

1.19 \ atm * 18.7 \ L = P_2 * 29.4 \ L

We are solving for the new pressure, so we must isolate the variable P_2. It is being multiplied by 29.4 liters. The inverse operation of multiplication is division. Divide both sides of the equation by 29.4 L.

\frac {1.19 \ atm * 18.7 \ L}{29.4 \ L} =\frac{ P_2 * 29.4 \ L}{29.4 \ L}

\frac {1.19 \ atm * 18.7 \ L}{29.4 \ L} =P_2

The units of liters cancel.

\frac {1.19 \ atm * 18.7 }{29.4 } =P_2

\frac {22.253}{29.4 } \ atm = P_2

0.7569047619 \ atm =P_2

The original measurements all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 9 in the ten-thousandth place to the right of this place tells us to round the 6 up to a 7.

0.757 \ atm \approx P_2

The pressure of the gas sample is approximately <u>0.757 atmospheres.</u>

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