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3 years ago

Complex organisms require a large number of cells that

2 answers:

Volgvan3 years ago

6 0

Answer: Thus, they ensure that the cells receive nutrients and can eliminate waste, thereby enabling them to survive and function properly.

Explanation: yes.

Romashka [77]3 years ago

3 0

Yes.
That’s the answer to your question

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Trial 12 1.Volume of Acid used ( mL) 2020 2.Molarity of acid used (M) 0.0160.016 3.Initial volume reading of base buret 018.6 4.

arsen [322]

Answer:

$V_B = 18.3mL$ -- Volume of base used

$M_B = 0.0175M$ --- Molarity of base

Explanation:

Given

$V_A = 20mL$ -- Volume of acid used

$V_B_1 = 18.6mL$ --- Buret Initial reading

$V_B_2 = 36.9mL$ --- Buret Final reading

$M_A = 0.016M$ --- Molarity of the acid

Solving (a): Volume of base used (VB)

This is calculated by subtracting the initial reading from the final reading of the base buret.

i.e.

$V_B = V_B_2 - V_B_1$

$V_B = 36.9mL - 18.6mL$

$V_B = 18.3mL$

Solving (b): Molarity base (MB)

This is calculated using:

$M_A * V_A = M_B * V_B$

Make MB the subject

$M_B = \frac{M_A * V_A }{V_B}$

This gives:

$M_B = \frac{0.016M *20mL}{18.3mL}$

$M_B = \frac{0.016M *20}{18.3}$

$M_B = \frac{0.32M}{18.3}$

$M_B = 0.0175M$

Solving (c): There is no such thing as average molarity

5 0

3 years ago

Organic chemistry please help

lord [1]

The answer to this is C. III & IV

3 0

4 years ago

This is my question.............

bazaltina [42]

Answer:

-3.15 °Celsius I think.

8 0

3 years ago

Read 2 more answers

4. Calculate the final concentration if water is added to 1.5 L of a 12 M

konstantin123 [22]

Answer:

Explanation:

(Molarity x Volume)concentrated soln = (Molarity x Volume)diluted doln

Molarity dilute soln = [(M x V)conc/V (dilute)] = 1.5L x 12M / 3.0L = 6M final dilute soln

6 0

4 years ago

Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 c

Inessa [10]

Answer: The mass of $AlCl_3$ produced is 889.38 g

Explanation:

We are given:

Moles of Al = 8 mol

Moles of $Cl_2$ = 10 mol

For the given chemical reaction:

$2Al+3Cl_2\rightarrow 2AlCl_3$

By stoichiometry of the reaction:

If 3 moles of chlorine gas reacts with 2 moles of Al

So, 10 moles of chlorine gas will react with = $\frac{2}{3}\times 10=6.67mol$ of Al

As the given amount of Al is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, chlorine gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 3 moles of $Cl_2$ produces 2 mole of $AlCl_3$

So, 10 moles of $Cl_2$ will produce = $\frac{2}{3}\times 10=6.67mol$ of $AlCl_3$

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We know, molar mass of $AlCl_3$ = 133.34 g/mol

Putting values in above equation, we get:

$\text{Mass of }AlCl_3=(6.67mol\times 133.34g/mol)=889.38g$

Hence, the mass of $AlCl_3$ produced is 889.38 g

4 0

3 years ago