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1 year ago

What happens when the density of a medium increases?

Chemistry

1 answer:

muminat1 year ago

8 0

I believe it’s Refractive index

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Calculate the total energy required to break all the bonds in:

Dennis_Churaev [7]

Explanation:

a. O+ 2=O2

b. C2+h2=CH4

c. C2+H6= C2H6

7 0

2 years ago

A cyclopropane-oxygen mixture is used as an anesthetic. If the partial pressure of cyclopropane, n the mixture is 330 mm Hg and

Thepotemich [5.8K]

b is the correct answer

4 0

3 years ago

Read 2 more answers

Find the potentials of the following electrochemical cell:

melomori [17]

Answer: 0.18 V

Explanation:-

$Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0_(Cd^{2+}/Cd)$ =-0.40V[/tex]

$E^0_(Ni^{2+}/Ni)$ =-0.24V[/tex]

$Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}$

$E^0=-0.24-(-0.40)=0.16V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}$

where,

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

$E^o_{cell}$ = standard electrode potential = 0.16 V

$E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}$

$E_{cell}=0.18V$

Thus the potential of the following electrochemical cell is 0.18 V.

6 0

2 years ago

g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4

kykrilka [37]

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

$Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2$

Stoichiometrically, 1 mole of $Na_2C_2O_4$ reacts with 2 moles of $HCl$ for a complete neutralization reaction.

Recall that: mole = $\frac{mass}{molar mass}$

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole $Na_2C_2O_4$ requires 2 moles HCl, then 0.0041 mole will require:

0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

7 0

3 years ago

"Infant Joy" is a conversation between __.

ElenaW [278]

The answer to this is C.

6 0

2 years ago