All categories

3 years ago

If you were allowed to do a science project and it was about the nervous system, why would you do that topic

Chemistry

2 answers:

Ket [755]3 years ago

4 0

I wouldn't because its too hard

Katarina [22]3 years ago

3 0

I would do the nervous system because nerves are what makes you feel everything around you and in order for you to be able to react you need the nervous system why? The nervous system is the highway along which your brain sends and receives information and feelings about what is happening in the body and around it on the outside. This highway is made up of billions of nerve cells, or neurons which join together to make nerves. This is very important for the human to be able to live!

You might be interested in

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall r

Sergeu [11.5K]

Answer:
5 moles of electrons

Explanation:

The balance equation is as follow,

 5 Ag⁺ + Mn⁺² + 4 H₂O → 5 Ag + MnO₄⁻ + 8 H⁺

Reduction of Ag:

Ag⁺ + 1 e⁻ → Ag
Or,
5 Ag⁺ + 5 e⁻ → 5 Ag

Oxidation of Mn:

Mn⁺² → MnO₄⁻ + 5 e⁻

Result:
Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².

4 0

3 years ago

2) which statement describes ionic bonding?

garik1379 [7]

Answer

odpowiedzi d

Explanation

taka jest reguła

7 0

3 years ago

Which of these is the best example of translational motion?

Dima020 [189]

The answer would b B, a leaf blowing across a field.
hope this helps.

6 0

3 years ago

Read 2 more answers

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

8 0

1 year ago

In comparison to an atom of F-19 in the ground state, an atom of C-12 in the ground state has(1) three fewer neutrons

mel-nik [20]

(2) three fewer valence electrons is your answer.

6 0

3 years ago