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Nat2105 [25]
4 years ago
12

Rhizobium:

Chemistry
2 answers:
Studentka2010 [4]4 years ago
7 0

Rhizobium: are nitrogen-fixing bacteria that live in the nodules of legumes which convert nitrogen in soil into nitrogen compounds. so the answer is all the above i hope this helps

Lapatulllka [165]4 years ago
4 0

Answer:

All of the above

Explanation:

Rhizobium is an important diazotrophic bacteria in nitrogen cycle. They are involved in a symbiotic relationship with leguminous plants; they assist in nitrogen fixation in the roots of these plant by taking up nitrogen from the atmosphere and passing it into the plants (in the form of ammonia or amino acids) which allows the plant to grow in soil with low nitrogen. The plant in exchange provide organic acids for the bacteria which it uses for carbon and energy source.

From the above, we can deduce that

(a) rhizobium are nitrogen fixing bacteria

(b) they live in the nodules of legumes (hence the symbiotic relationship)

(c) convert nitrogen in soil into nitrogen compounds like ammonia or amino acids

hence, the correct option is all of the above

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A gas has an initial volume of 52.3 L at 273 Kelvin. What is its temperature when the volume reaches 145.7 L?
erastovalidia [21]

Answer:

\boxed {\boxed {\sf 761 \ K}}

Explanation:

We are asked to find the new temperature of a gas after a change in volume. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume is initially 52.3 liters at a temperature of 273 Kelvin.

\frac {52.3 \ L}{273 \ K}= \frac{V_2}{T_2}

The volume reaches 145.7 liters at an unknown temperature.

\frac {52.3 \ L}{273 \ K}= \frac{145.7 \ L }{T_2}

We are solving for the new temperature, so we must isolate the variable T₂.  Cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

52.3 \ L * T_2 = 273 \ K * 145.7 \ L

Now the variable is being multiplied by 52.3 liters. The inverse of multiplication is division. Divide both sides by 52.3 L.

\frac {52.3 \ L * T_2 }{52.3 \ L}=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}

T_2=\frac{ 273 \ K * 145.7 \ L}{52.3 \ L}

The units of liters cancel.

T_2=\frac{ 273 \ K * 145.7 }{52.3 }

T_2 = 760.5372849 \ K

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we found, that is the ones place. The 5 in the tenths place tells us to round the 0 up to a 1.

T_2 \approx 761 \ K

When the volume reaches 145.7 liters, the temperature is <u>761 Kelvin.</u>

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3 years ago
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