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sweet [91]
3 years ago
12

If you have 2.00 mol of water, how many molecules of water is that?

Chemistry
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer: 1.204428152x10^24 molecules

Explanation:

1 mole of anything = 6.02214076x10^23 atoms/molecules/ions

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Most of the water a vascular plant takes in ______. Goes back into the soil Evaporates in the process of transpiration Is used t
MatroZZZ [7]

Answer:

EVAPORATES IN THE PROCESS OF TRANSPIRATION

Explanation:

Plants takes in water from the soul via it's root system. Small amount of the water are used for growth and metabolism of the plants. Some are used in the process of photosynthesis. The bulk of the remaining are lost through transpiration. This takes place in the stomata of the plants. The stomata are numerous on the leaf surfaces and are guarded by cells which allow for opening and closing of the stomata. Transpiration helps the plants to remove CO2 produced by photosynthesis, cools the plant and also changes osmotic pressure of the plant which allows for the flow of nutrients and water from the root to the leaves, flowers, fruits.

7 0
3 years ago
One of the main ideas of the Quantum Theory is that:
gladu [14]
The energy of an electron can be change only by set amounts.

The electrons in an energy level of an atom can only move to a higher or lower energy level only if it absorbs or emits a certain amount of energy. If the electron does not have enough energy to move to the next energy level then it will never move.
8 0
3 years ago
Read 2 more answers
ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved.
SSSSS [86.1K]

Answer:

Therefore, the number of grams of salt in the tank at time t is A(t) = 150-140 e^{-\frac{t}{30} }

Explanation:

Given:

Tank A contain V_{1} = 150 lit

Rate \alpha = 5 \frac{L}{min}

Dissolved salt A = 10 gm

Salt pumped in one minute is 4 \frac{L}{min}

Salt pumped out is \frac{5L}{150L} = \frac{1}{30} of initial amount added salt.

To find A(t)

  \frac{dA}{dt}  = Rate _{in} - Rate _{out}

  A' = 5 - \frac{A}{30}

  A' + \frac{A}{30} = 5

Solving above equation,

  I .F   = e^{\int\limits {p} \, dt }

   y = e^{\int\limits {\frac{1}{30} } \, dt }

   y = e^{\frac{t}{30} }

(Ae^{\frac{t}{30} }  )' = 5 e^{\frac{t}{30} } + c

Integrating on both side,

Ae^{\frac{t}{30} } =  5 \times 30 e^{\frac{t}{30} } +c

Add e^{-\frac{t}{30} } on above equation,

 A = 150 + ce^{-\frac{t}{30} }

Here given in question,A(t=0) = 10

 10 =150 +c

   c = -140

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

 A(t) = 150-140 e^{-\frac{t}{30} }

Therefore, the number of grams of salt in the tank at time t is A(t) = 150-140 e^{-\frac{t}{30} }

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3 years ago
Calculate [H+] if [OH-] = 3.7 X 10-5 M
barxatty [35]

Answer:

turtle

Explanation:

i like turtles

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4 years ago
HELPPPPP<br><br> A: <br> B: <br> C: <br> D: <br> E:
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