All categories

3 years ago

Which equation represents sublimation?

Chemistry

2 answers:

Zanzabum3 years ago

6 0

The right answer is 2 because it changed from solid to gas.

Alexxx [7]3 years ago

3 0

The right answer is 2

You might be interested in

How many atoms are in 8 grams of helium

Mrrafil [7]

12 x 10^23 atoms, i hope this helps

8 0

3 years ago

In the compound cah2 calcium has an oxidation number of 2+ and hydrogen has an oxidation number of

sesenic [268]

The oxidation number of H is -1.

Sum of the oxidation numbers in each element = charge of the complex

CaH₂ has 1 Ca atom and 2H atoms. The charge of the complex is zero. Let’s say Oxidation number of H is "a".

Then,

(+2) + 2 x a = 0

+2 + 2a = 0

2a = -2

a = -1

Hence, the oxidation number of Hydrogen atom in CaH₂ is -1

7 0

3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

Calculate the mass in 4.05*10^22 molecules of calcium phosphate

Marizza181 [45]

Answer:

m = 20.9 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:

$m=4.05x10^{22}molecules*\frac{1mol}{6.022x10^{23}}*\frac{310.18g}{1mol}\\\\m=20.9g$

Regards!

8 0

3 years ago

A factor that does no change in an experiment is the.

kvasek [131]

Answer:

A. Controlled variable

Explanation:

a controlled variable or a constant variable is a variable that doesnt change during an experiment

5 0

3 years ago