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Licemer1 [7]
2 years ago
8

Two reasons that a plant from a rain forest would not survive in a desert home?

Chemistry
2 answers:
arsen [322]2 years ago
7 0

okay, the answer should be simple and easy, plants from a rainforest would not survive in a desert because the plants in the rainforest do get water, sunlight and air whereas the plants in the desert have Sunlight which is very hot. so if a plant in the rainforest was transferred to the desert it will hard to survive and it will eventually die because they will not get water that will be needed and the plant can get sunburns with the too hot temperature.

ss7ja [257]2 years ago
4 0

Answer: A planet from a rain forest would not survive in a desert home, for the fact they depend on water and as these live from water, without it, this leads to dehydration and with loss of water with plants, nothing is able to survive when it comes to a rain forest plant being in a desert home. Another reason is the high trees and leaves that are providing the plants enough sunlight or shade to grow, it guarantees to help keep the temperature normal. Being in a deserted area would mean that there would be a temperature change, something the plant is not used to. Without that needed shade and avoiding the scorching sun, they will die out from how different the temperature is, and how hot it is.

Explanation: I hope this helped you.

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7. Calculate the amount of energy required heat 100.g to <br> H2O(s) changes to H2O(l) at 0°C
Brrunno [24]

Answer:

33300J

Explanation:

Given parameters:

Mass of ice  = 100g

Unknown:

Amount of energy = ?

Solution:

This is a phase change process from solid to liquid. In this case, the latent heat of melting of ice is 3.33 x 10⁵ J/kg.

 So;

            H  = mL

m is the mass

L is the latent heat of melting ice

  Now, insert the parameters and solve;

            H = mL

 mass from gram to kilogram;

            100g gives 0.1kg

  H = 0.1 x 3.33 x 10⁵ = 33300J

8 0
2 years ago
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place
kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

7 0
3 years ago
1. What is the difference between accidental discoveries and scientific investigations?
Alenkasestr [34]

Answer:

the discovery will always need to be tested more where if you're investigating you should already be in a more controlled environment

4 0
2 years ago
Hey can someone help me figure out what ph of ajax cleaner and what color it changes to?
serious [3.7K]
I just know the ph is between 7 and 8
8 0
3 years ago
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Helga [31]

Answer:

<em> 14, 508J/K</em>

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

<em>m = mass of substance (g) = 0.1184g</em>

1 mole of Mg - 24g

<em>n</em> moles - 0.1184g

<em>n = 0.0049 moles.</em>

Also, q = m × c × ΔT

<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>

<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>

<em>= 14, 508 J/K/kg</em>

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>

<em>= 1,7117.7472 J °C-1 g-1</em>

<em />

<em>∴ ΔHrxn = q/n</em>

<em>=1,7117.7472  ÷ 0.1184 </em>

<em>= 14, 508J/K</em>

6 0
3 years ago
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