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4 years ago

Plant cells have a large central vacuole, which animal cells lack. What function does this organelle perform?

Chemistry

2 answers:

Lapatulllka [165]4 years ago

5 0

The central vacuole stores materials, wastes, and helps give the plant structure and support.

Hope this helps!

ExtremeBDS [4]4 years ago

3 0

Answer:

A Plant cell contains a large vacuole which is membrane-bound organelle specialized to perform several unique functions in the plant cells like:

1. It is used by the plant to store the waste material of the cell.

2. Its major function is to store the water which provides shape and rigidity to the plant cell.

3. It maintains the rigidity as it helps maintain the turgor pressure of the plant cell.

4. It also stores the ions of the cell and thus controls the pH of the cell.

Thus, a plant vacuole serves many functions to the plant cell.

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In the molecule ClF3, chlorine makes three covalent bonds. Therefore, three of its seven valence electrons need to be unpaired.

Kisachek [45]

Answer:

Explanation:

Chlorine has electronic configuration of 2 , 8 , 7

In n = 3 there are 7 electrons out of which 2 are in s , and 5 are in p . But out of 5 electrons in p , one electron jumps into d orbital . so the electronic configuration becomes as follows

$3s^23p_x^23p_y^23p_z^1$ = 7

$3s^23p_x^23p_y^13p_z^13d_{xy}^1$

These orbitals like sp³d hybridise to form 7 degenerate orbitals out of which 2 orbitals contain electrons in pairs and rest three are singly occupied by electrons.( unpaired electrons )

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3 years ago

Why does air at 75% relative humidity and 30°C make your skinfeel stickier than air at 75% relative humidity and 10°C

ahrayia [7]

Here are the choices:

Warm air rises while cold air falls.

The warmer air contains less water vapor per unit of volume.

The warmer air contains more water vapor per unit of volume.

Moist air has lower density than dry air does.

The best answer is: <em>The warmer air contains more water vapor per unit of volume. </em>

8 0

3 years ago

Read 2 more answers

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gregori [183]

Need more to the question

5 0

4 years ago

A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp

MArishka [77]

Answer:

V = 11.2L are produced

Explanation:

... <em>Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..</em>

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

<em>Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.</em>

<em />

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

<em>Moles Al-Molar mass: 26.98g/mol-:</em>

27g * (1mol / 26.98g) = 1mol of Al

<em>Moles HCl:</em>

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and <em>HCl is limiting reactant.</em>

<em />

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

<em>Where V is volume</em>

<em>n are moles: 0.5mol</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 273.15K</em>

<em>P is pressure: 1atm.</em>

<em />

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>

3 0

3 years ago

Calculate the [OH-] and the pH of a solution with an [H+= .00083 M at 25 degrees

sesenic [268]

<h3>Answer:</h3>

1.2 × 10^-11 M

<h3>Explanation:</h3>

<u>We are given;</u>

Concentration of Hydrogen ions [H⁺] as 0.00083 M

<u>We are required to calculate the concentration of [OH⁻]</u>

We know that;

pH = - log [H⁺]

POH = -log[OH⁻]

Also, pH + pOH = 14

With the concentration of H⁺ we can calculate the pH

pH = -log 0.00083 M

= 3.08

But; pH + pOH = 14

Therefore; pOH = 14 - pH

= 14 - 3.08

= 10.92

But, pOH = -log[OH⁻]

Therefore; [OH] = -Antilog(pOH)

Hence; [OH⁻] = Antilog -10.92

= 1.2 × 10^-11 M

Therefore, [OH⁻] is 1.2 × 10^-11 M

6 0

3 years ago