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2 years ago

The diagram below shows a certain method of grazing animals

Biology

2 answers:

WITCHER [35]2 years ago

5 0

Answer:

In agriculture, rotational grazing, as opposed to continuous grazing, describes many systems of pasturing, whereby livestock are moved to portions of the pasture, called paddocks, while the other portions rest.

Explanation:

hehe ....

Naya [18.7K]2 years ago

5 0

Hey ,

Can you please attached the diagram ...

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3 years ago

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3 years ago

An 80-year-old client with a history of coronary artery disease is admitted to the hospital for observation after a fall. during

lara31 [8.8K]

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The client's paroxysmal dyspnea was probably caused by sleeping in bed with the legs at the level of the heart; this position increases venous return from dependent body areas, increasing the intravascular volume. Sitting up and leaning forward while keeping the legs dependent slows venous return as well as increases thoracic capacity. Although the contour position elevates the client's head, it does not place the legs in a dependent enough position to substantially decrease venous return. The recumbent position is contraindicated. Venous return increases when the lower extremities are at the level of the heart. Also, the pressure of the abdominal organs against the diaphragm decreases thoracic capacity. The Trendelenburg position is contraindicated. Venous return increases when the lower extremities are higher than the level of the heart. Also, the pressure of the abdominal organs against the diaphragm decreases thoracic capacity.

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3 years ago

How many grams of urea (non volatile, non electrolyte) must be added to 1250. g water to give a solution with a vapor pressure l

Lady bird [3.3K]

Answer:

$\large\boxed{\large\boxed{107g}}$

Explanation:

The vapor pressure lowering is a colligative property and it follows Raoult's law.

The vapor pressure lowering of a solvent in a solution, ΔP, is equal to the mole fraction of the solute, Xsolute, multiplied by the vapor pressure of the pure solvent, P°.

$\Delta P=X_{solute}\times P^0$

ΔP and P⁰ are given:

ΔP = 1.39mmHg

P⁰ = 55.32mmHg

1. Mole fraction of solute

Thus, you can calculate Xsolute:

$1.39mmHg=X_{solute}\times 55.32mmHg\\\\X_{solute}=0.02513$

2. Moles of solute

Now you can calculate the number of moles of water and the number of moles of solute.

Number of moles of water = mass in grams / molar mass

Number of moles of water = 1,250g/18.015 (g/mol) = 69.3866mol

Xsolvent = 1 - Xsolute = 1 - 0.02513 = 0.97487

Xsolvent = moles of solvent / moles of solution

0.97487 = 69.3866mol / moles of solution

moles of solution = 71.1752mol

moles of solute = moles of solution - moles of solvent = 71.1752mol - 69.3866mol = 1.7886mol

3. Mass of urea

Formula:

Mass = number of moles × molar mass

Mass = 1.7886 mol × 60.06 g/mol = 107.42 g

You must round to 3 significant figures: 107 g

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3 years ago