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3 years ago

Nitrogen fixing bacteria in soil turns nitrogen gas into

1 answer:

4 0

Nitrogen gas is converted to nitrate compounds by nitrogen-fixing bacteria in soil turns nitrogen gas into root nodules. Nitrogen is the most commonly limiting nutrient in plants. Legumes use nitrogen fixing bacteria, specifically symbiotic rhizobia bacteria, within their root nodules to counter the limitation.

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Which instrument below is used to measure an intensive property?

Andrei [34K]

An intensive property is the physical characteristics that have an independent magnitude. The thermometer can be used to measure the temperature. Thus, option C is correct.

<h3>What is an intensive property?</h3>

An intensive property has been constituted of the parameters that are not dependent on the size and the mass of the sample. Density, pressure, and temperature are some intensive properties.

The first image shows a weighing balance, the second shows a volumetric cylinder, and the fourth shows a ruler used to measure mass, volume, and length respectively, which are extensive properties.

Therefore, option C. thermometer measures temperature, which is an intensive property.

Learn more about the intensive property here:

brainly.com/question/17323212

#SPJ1

4 0

1 year ago

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T

taurus [48]

Answer:

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

Adding solute to water causes the depression of the freezing point.

We have the relation:

ΔTf = Kf.m,

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

Molality is the no. of moles of solute per kg of the solution.

no. of moles of solute (glucose) = mass/molar mass = (8.44 g)/(180.156 g/mol) = 0.04685 mol.

∴ molality (m) = no. of moles of solute/kg of solvent = (0.04685 mol)/(2.11 kg) = 0.0222 m.

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

∴ The freezing point of the solution = the freezing point of water - ΔTf = 0.0°C - 0.0413°C = - 0.0413°C ≅ - 0.041°C (nearest thousands).

6 0

3 years ago

Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/

Ivenika [448]

Answer:

The answer to your question is: C₃H₃O This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

44 g -------------------- 12 g

3.2007 g ------------ x

x = (3.2007 x 12) / 44

x = 0.8729 g of Carbon

12 g of C -------------- 1 mol

0.8729 g -------------- x

x = (0.8729 x 1) / 12

x = 0.0727 mol of Carbon

For Hydrogen

18 g ---------------------- 1 g

1.3102 g ------------------- x

x = (1.3102 x 1) / 18

x = 0.0727 g of Hydrogen

1 g ------------------------ 1 mol

0.0727g ---------------- x

x = (0.0727 x 1)/1

x = 0.0727 mol of Hydrogen

For oxygen

g of Oxygen = g of sample - g of Carbon - g of hydrogen

g of Oxygen = 1.3109 - 0.8709 - 0.0727

g of Oxygen = 0.3673

16 g of Oxygen ------------- 1 mol of O

0.3673 g --------------------- x

x = (0.3673 x 1)/ 16

x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon 0.0727 / 0.023 = 3.1 ≈ 3

Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3

Oxygen 0.0230 / 0.023 = 1

C₃H₃O

8 0

3 years ago

Read 2 more answers

The electron affinity of Bi is -94.6 kJ and the value for At is -280 kJ. What is the approximate electron affinity for Po?

lys-0071 [83]

Answer:

48 KJ mol-1

Explanation:

3 0

2 years ago

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

$0.050mol \times \frac{149.09 g}{mol} = 7.5 g$

6 0

2 years ago