Ask question

Ask question

All categories

2 years ago

11

What is electromagnetic radiation?

1 answer:

slega [8]2 years ago

5 0

That’s what I’m trying to figure out

You might be interested in

Take it before its gone

MaRussiya [10]

Hello my friend thank you

8 0

3 years ago

Bases turn litmus paper what color?<br><br> a) red<br> b) blue<br> c) yellow<br> d) pink

yKpoI14uk [10]

Answer:
blue
.......
.......

5 0

3 years ago

Read 2 more answers

What is the mass of 1.71 ✕ 1023 molecules of h2so4?

Anuta_ua [19.1K]

The equation for calculating a mass is as follows:

m=n×M

Molar mass (M) we can determine from Ar that can read in a periodical table, and a number of moles we can calculate from the available date for N:

n(H2SO4)=N/NA

n(H2SO4)= 1.7×10²³ / 6 × 10²³

n(H2SO4)= 0.3 mole

Now we can calculate a mass of H2SO4:

m(H2SO4) = n×M = 0.3 × 98 = 27.8 g

7 0

3 years ago

Iron chloride + sodium hydroxide

patriot [66]

Answer:

Sodium hydroxide reacts with iron(III) chloride to produce iron(III) hydroxide and sodium chloride.

Hope this helps plz hit the crown :D

8 0

2 years ago

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi

avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction : $Ni^{2+}+2e^-\rightarrow Ni$

The expression for reaction quotient will be :

$Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(0.0141)}{(0.00104)}=13.6$

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

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

$E^o_{cell}$ = standard electrode potential of the cell = 0.51 V

$E_{cell}$ = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

$E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)$

$E_{cell}=0.47V$

Therefore, the actual cell potential of the cell is 0.47 V

4 0

3 years ago