3 years ago

Identify whather the following are elements, compounds and mixture.

Chemistry

1 answer:

astra-53 [7]3 years ago

6 0

1 is element 2 is element 3 is compound 4 is element 5 is nothing in and water is a compound

You might be interested in

Classify the following compounds as binary ionic, ternary ionic, or molecular.

Alborosie

Answer:bro u already have an answer why are you asking?

Explanation:

3 0

3 years ago

Molecular formula of C4H5

BARSIC [14]

Answer:

Cyclopropylmethylene

7 0

3 years ago

Solute: Will give brainliest

gtnhenbr [62]

Answer:

58.45 g of NaCI

Explanation:

hope this helps ya

have a great day

brainliest if its right?

4 0

2 years ago

The combining of two atomic nuclei to produce one larger nucleus is called

Marta_Voda [28]

<span>The combining of two atomic nuclei to produce one larger nucleus is called nuclear fusion. </span>

5 0

2 years ago

A concentration cell is one in which both the anode and cathode are the same but with different concentrations. Calculate the ce

kolezko [41]

Answer: 0.029 V

Explanation:

For the given chemical reaction :

$Zn^{2+}(0.10M)(aq)+Zn(s)\rightarrow Zn^2+{0.010M)(aq)+Zn(s)$

Using Nernst equation :

$E_{cell}=E^0_{cell}-\frac{2.303RT}{nF}\log\frac{\text {anodic ion concentration}}{\text {cathodic ion concentration}}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 298 K

n = number of electrons in oxidation-reduction reaction = 2

$E^0_{cell}$ = standard electrode potential of the cell = 0 (as both metals are same )

$E_{cell}$ = emf of the cell = ?

$E_{cell}=0-\frac{2.303\times 8.314\times 298}{2\times 96500}\log\frac{0.010}{0.10}$

$E_{cell}=0.029V$

Thus the cell potential will be 0.029 V

4 0

2 years ago