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Julli [10]
3 years ago
12

Other than bacteria, which factor leads to nitrogen fixation?

Chemistry
2 answers:
alina1380 [7]3 years ago
4 0
Lightning rain can lead to nitrogen fixation other than bacteria
bezimeni [28]3 years ago
3 0
  the answer is lightning<span />
You might be interested in
Hydrogen is an element with two naturally occurring isotopes: 2 H and 3 H. This means that 2 H, which has a mass number of 2, ha
Dmitry_Shevchenko [17]

Answer:

TRUE.

Explanation:

Mass Number is the sum of protons and neutrons present in the nucleus of an atom. Isotopy is a phenonmenon that occurs when atoms of same elements have different mass number (Number of neutrons).

2H isotope has 1 proton and 1 neutron.

3H isotope has 1 proton and 2 neutrons.

This meeans 2H isotope has fewer neutrons when compared to the 3H isotope. The correct option is TRUE.

8 0
3 years ago
A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th
Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.  

The mass percent of lithium hydroxide can be calculated with the following equation:  

\% = \frac{m_{LiOH}}{m_{t}} \times 100    (1)

Where:

m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g   (2)  

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.    

\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}

0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}

\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    (3)                                        

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:  

\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    

\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol              

Solving for m_{LiOH}, we have:

m_{LiOH} = 0.082 g

Hence, the percent lithium hydroxide is (eq 1):

\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%  

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

  • brainly.com/question/6992535?referrer=searchResults
  • brainly.com/question/5840377?referrer=searchResults

I hope it helps you!                        

5 0
2 years ago
What is the pH of a solution with a 3.8 × 10−4 M hydronium ion concentration?
mamaluj [8]

Answer:

3.4

Explanation:

The pH scale is used to express the acidity or basicity of a solution.

  • If the pH < 7, the solution is acid.
  • If the pH = 7, the solution is neutral.
  • If the pH > 7, the solution is basic.

Given the hydronium ion concentration [H₃O⁺] = 3.8 × 10⁻⁴ M, we can calculate the pH using the following expression.

pH = -log [H₃O⁺]

pH = -log 3.8 × 10⁻⁴

pH = 3.4

This solution is acid.

8 0
3 years ago
Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) =&gt; CH 3OH(l) Using the following information: C(graphite) +
Alika [10]

Answer:

\Delta H for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

C(graphite)+O_{2}(g)\rightarrow CO_{2}(g) \Delta H_{1}= -393.5 kJ

2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l) \Delta H_{2}= (2\times -285.8)kJ

CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)  \Delta H_{3}= 726.4 kJ

---------------------------------------------------------------------------------------------------

C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)

\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ

4 0
4 years ago
Calculate the mass of carbon dioxide released when 185.000 g of copper carbonate [molar mass = 123.5 g/mol] is heated (assuming
Vinvika [58]

Answer:

The answer to your question is: 65.9 g released of CO2

Explanation:

MW CO2 = 44 g

MW CuCO3 = 123.5 g

CO2 released = ?

CuCO3 = 185 g

                            CuCO3     ⇒    CO2   +    CuO

                             123.5  -----------  44g

                             185 g -----------    x

                           x = (185 x 44) / 123.5

                           x = 65.9 g released of CO2

8 0
3 years ago
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