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1 month ago

How does vegetation slow and prevent sediment loss?

1 answer:

5 0

Vegetation cover serves as the most efficient and effective method to curb sediment loss. The roots of plants like grass interlink soil particles, aiding in erosion resistance, particularly against runoff water. Vegetation absorbs the force of raindrops, preventing soil particle detachment. Additionally, plants can lie flat resembling shingles on a roof, enabling runoff to travel over the soil rather than disturbing it. Tall, erect vegetation functions as a barrier against wind, diminishing its force so that it cannot dislodge soil particles from the ground surface.

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If 7.50 g of the unknown compound contained 0.250 mol of c and 0.500 mol of h, how many moles of oxygen, o, were in the sample?

lions [2927]

The result will be 100.60 because we are adding to the atmosphere, which connects to the lubricator, and then it turns, resulting in 100.60.

3 0

1 month ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

KiRa [2933]

To determine the specific heat capacity of the metal and assist in its identification, the heat absorbed by the calorimeter can be computed using: Energy = mass * specific heat capacity * temperature change Q = 250 * 1.035 * (11.08 - 10) Q = 279.45 cal/g. Next, we employ the same formula for the metal because the heat taken in by the calorimeter should equal the heat expelled by the metal. -279.45 = 50 * c * (11.08 - 45) [the minus sign indicates energy release] solving for c gives us 0.165. Therefore, the specific heat capacity of the metal amounts to 0.165 cal/g°C.

6 0

1 month ago

Read 2 more answers

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

KiRa [2933]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

1 month ago

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

castortr0y [3046]

Answer:

81°C.

Justification:

We can arrive at this conclusion using the formula:

Q = m.c.ΔT,

where Q denotes the heat lost by water (Q = - 1200 J).

m represents the mass of water (m = 20.0 g).

c indicates the specific heat of water (c = 4.186 J/g.°C).

ΔT signifies the difference between the starting temperature and the final temperature (ΔT = final T - initial T = final T - 95.0°C).

Given Q = m.c.ΔT

It follows that (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

Consequently, the correct answer is: 81°C.

7 0

1 month ago

An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.

KiRa [2933]

Response:

0.8853 mL

Clarification:

Initially, we convert 13 lb to kg, remembering that 1 lb = 0.454 kg:

13 lb * $\frac{0.454kg}{1lb}$ = 5.902 kg

Next, we determine the required mg of acetaminophen to administer, applying the recommended dosage and infant's weight:

15 mg/kg * 5.902 kg = 88.53 mg

Finally, we compute the necessary mL of suspension, utilizing its concentration:

88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mL

8 0

1 month ago