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4 years ago

Which characteristic makes legumes a good food source for food-insecure populations?

Chemistry

2 answers:

vova2212 [387]4 years ago

6 0

Legumes are much easier to grow than other plants, and are more adaptable.

Ugo [173]4 years ago

4 0

Answer:

Legumes have numerous advantages for health and are also very good for soils.

Explanation:

From a health point of view, according to the Food and Agriculture Organization of the UN, legumes are seeds that contain low sugar, low fat, high fiber, iron and antioxidants, all of which contribute to a better functioning of the human body. But also from an ecological perspective, this plants are great for restoring soils because their are the only group of plants capable of sequestering nitrogen from the atmosphere into the food webs. In food insecure populations usually one of the main problems is the poor state of soils, and also the need for high-nutrient aliments to fight diseases. That is why legumes are such a good source for them.

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Which choice describes the forms that energy takes when it starts in the Sun and ends in the sugars of plants?

Lapatulllka [165]

It goes from radiant to chemical.

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4 years ago

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How did Wilma contribute to change in America

Natasha_Volkova [10]

Answer:

She overcame her disabilities to compete in the 1956 Summer Olympic Games, and in 1960, she became the first American woman to win three gold medals in track and field at a single Olympics. Later in life, she formed the Wilma Rudolph Foundation to promote amateur athletics.

Explanation:

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3 years ago

This is due in five mintues please help

Fittoniya [83]

Answer:

Explanation:

Na, N, O, O, O

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3 years ago

Will a precipitate form if you mix 75.0 mL of a NaOH solution with pOH of 2.58 if it is mixed with 125.0 mL of a 0.0018 M MgCl2

r-ruslan [8.4K]

A precipiatate of mass 0.013 g is formed when you mix 75.0 mL of a NaOH solution with pOH of 2.58 and 125.0 mL of a 0.0018 M MgCl2 solution

<h3>What is stoichiometry?</h3>

In stoichiometry, calculations are made based on mass - mole or mole - volume relationships. First we must put down the balanced reaction equation; 2NaOH(aq) + MgCl2(aq) -----> Mg(OH)2(s) + 2NaCl(aq)

Now Ionically;

2OH^-(aq) + Mg^2+(aq) -----> Mg(OH)2(s)

Concentration of OH^- = Antilog (-2.58) = 2.6 * 10^-3 M

Number of moles of OH^- = 2.6 * 10^-3 M * 75/1000 = 0.000195 moles

Concentration of Mg^2+ = 0.0018 M

Number of moles of Mg^2+ =0.0018 M * 125/1000 = 0.000225 moles

Since;

1 mole of Mg^2+ reacts with 2 moles of OH^-

x moles of Mg^2+ reacts with 0.000195 moles of OH^-

x = 0.0000975 moles

Mg^2+ is the limiting reactant.

1 mole of Mg^2+ yields 1 mole of the precipitate

0.000225 moles of Mg^2+ yields 0.000225 moles of precipitate.

Hence, a precipitate is formed.

Learn more about precipitate: brainly.com/question/1770619?

5 0

3 years ago

⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a

STALIN [3.7K]

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

<u>For krypton</u>:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

$P x V= \frac{m}{MM} x R x T$

m/V= $\frac{P x MM}{R x T}$

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

= (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

= 2.81 g/L

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = $\frac{dx R x T}{P}$

= (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

= 63.99 g/mol ≅ 64 g/mol

<u>For the O₂</u>:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒ m = $\frac{P x V x MM}{R x T}$ = (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

= 15.29 g≅ 16 g

4 0

3 years ago

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