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3 years ago

11

Describe how three of the organelles work together within a plant cell.NEED HELP ASAP

1 answer:

7nadin3 [17]3 years ago

5 0

For cells to work and survive,they all carry specific tasks which leaves impact on the cell, in short terms all the cells are important and contribute for the cell to survive.

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How do three states of matter arise?

Evgen [1.6K]

The status of the state of matter of a material defines the phase. Pellet currently case 16 is defined; three of which solid, liquid and gas are referred to as conventional cases. There are a total of 16 state currently defined, these classic except; liquid crystal, amorphous solid, magnetic regularly, superconducting, superfluid, the Bose-Einstein condensation, Rydberg molecules, plasma (ionized gas), the quark-gluon plasma, degenerate matter, Superkate, the stringy liquid and SuperCam mine. Daily life in general, solid materials, include liquid or gaseous, but outside world, 99% of the material in the universe plasma (ionized gas) is in the form.

6 0

3 years ago

How many mL of .450 M HCL is needed to neutralize 30mL of .150 M Ba(OH)2

Aleks04 [339]

You need .556M HCL to neutralize that

7 0

2 years ago

PLEASE HURRY THIS IS TIMED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

AURORKA [14]

Answer:

if i remember correctly i beleive its A 1.8 x 10^24

but im not for sure also i think you forgot the 24

Explanation:

5 0

3 years ago

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

3 years ago

A container with 3.0 moles of gas has a volume of 60.0L with a temperature at 400.K what is the pressure

iris [78.8K]

Answer: P= 1.64 atm

Explanation: solution attached.

Use Ideal gas law

PV= nRT

Derive for P

P= nRT/V R= 0.08205 L.atm/mol.K

Substitute the values.

3 0

3 years ago