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kotegsom [21]
2 years ago
8

7. The root structures on peanut plants that restore soil nutrients and fix nitrogen​

Chemistry
1 answer:
timofeeve [1]2 years ago
3 0

Answer:

Like other legumes, peanut plants improve the soil by adding nitrogen, even as they grow tasty, nutritious nuts for this season's harvest. But peanuts need help as they take nitrogen from the air and "fix" it into the soil via their root systems.

Explanation:

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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

3 0
3 years ago
A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?
SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar  mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M =  0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K /  1 atm

M = 20.5 g/mol

8 0
2 years ago
How many liters of octane are in a shipping container that contains 45 moles?
Andrew [12]
D. 1,008 liters because you are looking for liters from a calculation of moles. Recognizing that you can do STP (22.4L) you multiply this number by 45 moles and it is 1,008 liters
3 0
2 years ago
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A(n) ____________________ is a negatively charged subatomic particle.
gtnhenbr [62]
An electron is a negatively charged subatomic particle, whereas a proton is positively charged, and a neutron has no charge. 
8 0
3 years ago
C -14 and N-14 both have same mass number yet they are different elements. Explain
KiRa [710]

Explanation:

hdy hey hey hegzyv vu g g ycgx

7 0
3 years ago
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