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Leviafan [203]
2 years ago
10

How do the Carnivorous plants survive without soil?

Chemistry
1 answer:
Mkey [24]2 years ago
5 0
Answer:

Carnivorous plants are easy to grow, if you follow a few, simple rules.

Wet all of the time.
Mineral-free water.
Mineral-free soil.
Lots of light.


Wet all of the time.
Carnivorous plants are native to bogs and similar nutrient-poor habitats. As a consequence, the plants live in conditions that are constantly damp. To grow healthy carnivorous plants, it is important to duplicate their habitat as closely as possible. Keep the soil wet or at least damp all of the time. The easiest way to do this is use the tray method. Set the pots in a tray or saucer, and keep water in it at all times. Pitcher plants can grow in soggy soil with the water level in the saucer as deep as 1/2 the pot, but most carnivorous plants prefer damp to wet soil, so keep the water at about 1/4 inch and refill as soon as it is nearly gone. Water from below, by adding water to the tray, rather than watering the plant. This will avoid washing away the sticky muscilage of the sundews and butterworts and keep from closing the flytraps with a false alarm.


Mineral-free water.
Always use mineral-free water with your carnivorous plants, such as rainwater or distilled water. Try keeping a bucket near the downspout to collect rainwater. Distilled water can be purchased at the grocery store, but avoid bottled drinking water. There are simply too many minerals in it. The condensation line from an air conditioner or heat pump is another source of mineral-free water. Reverse-osmosis water is fine to use. Carnivorous plants grow in nutrient poor soils. The minerals from tap water can “over-fertilize” and “burn out” the plants. In a pinch, tap water will work for a short while, but flush out the minerals with generous portions of rainwater, when it is available.


Mineral-free soil.
The nutrient poor soils to which the carnivorous plants have adapted are often rich in peat and sand. This can be duplicated with a soil mixture of sphagnum peat moss and horticultural sand. Be sure to check the peat label for sphagnum moss. Other types will not work well. The sand should be clean and washed. Play box sand is great, and so is horticultural sand. Avoid “contractor’s sand” which will contain fine dust, silt, clay and other minerals. Never use beach sand or limestone based sand. The salt content will harm the plants. The ratio of the mix is not critical, 1 part peat with 1 part sand works well for most carnivorous plants. Flytraps prefer a bit more sand, and nepenthes prefer much more peat. Use plastic pots, as terra cotta pots will leach out minerals over time and stress your plants.

Explanation:


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If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced
ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • P₄: 1 mole
  • H₂: 6 moles
  • PH₃:4 moles

Being the molar mass of the compounds:

  • P₄: 124 g/mole
  • H₂: 2 g/mole
  • PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

  • P₄: 1 mole* 124 g/mole= 124 g
  • H₂: 6 mole* 2 g/mole= 12 g
  • PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

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