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Nataly_w [17]
3 years ago
14

A lithium atom had 3 protons 3 neutrons and 3 electrons what is the overall charge on its atom

Chemistry
2 answers:
irga5000 [103]3 years ago
8 0
Wait what supject is this chemistry? Unknown I'll help you after
Ainat [17]3 years ago
8 0
The overall charge on the atom is zero

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I have a question on my Chem homework. It's asking "How many molecules of N2O4 are in 76.3 g N2O4? The molar mass of N2O4 is 92.
evablogger [386]
Here, Molar mass of N2O4 = 92.02 g/mol
So, Number of molecules in 1 mole (92.02 g) = 6.023 * 10²³
Number of molecules in 76.3 g = 6.023 * 10²³ / 92.02  * 76.3
= 6.023 * 10²³ / 1.20
= 5 * 10²³ 

In short, Your Answer would be approx. 5 * 10²³ molecules

Hope this helps!
7 0
4 years ago
What mass of iron is formed when 10 grams of carbon react with 80 grams of iron iii oxide?
yKpoI14uk [10]

Answer:

55.85 grams of Fe is formed.

Explanation:

Identify the reaction:

2Fe₂O₃  +  3C  →  4Fe  +  3CO₂

Identify the limiting reactant, previously determine the mol of each reactant

(mass / molar mass)

10 g / 12 g/m = 0.83 moles C

80 g / 159.7 g /m = 0.500 moles Fe₂O₃

2 moles of oxide need 3 moles of C, to react

0.5 moles of oxide, will need ( 0.5  . 3)/ 2 = 0.751 mol

I have 0.83 moles of C, so C is the excess.

The limiting is the oxide.

3 mol of C need 2 mol of oxide to react

0.83 mol of C, will need (0.83  . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)

Ratio is 2:4 (double)

If I have 0.5 moles of oxide, I will produce the double, in the reaction.

1 mol of Fe, will be produce so its mass is 55.85 g

5 0
3 years ago
Leopards and cheetahs look very similar, since both have spotted coats, but the two species of big cats are actually quite diffe
Lisa [10]
The answer is letter A i think
3 0
3 years ago
If a gas has a volume of 350 mL at 780 mm Hg, what pressure will be needed to cause its volume
makkiz [27]

The required pressure of the gas is 546 mmHg.

<h3>What is the relation between volume and pressure?</h3>

Relation between the volume and pressure of gas is explained through the ideal gas equation PV = nRT, and for the question required equation is:

P₁V₁ = P₂V₂, where

P₁ & V₁ are the initial pressure and volume.

P₂ & V₂ are the final pressure and volume.

On putting values from the question to the equation, we get

P₂ = (780)(350) / (500) =

P₂ = 546 mmHg

Hence required pressure is 546 mmHg.

To know more about ideal gas equation, visit the below link:

brainly.com/question/24236411

#SPJ1

5 0
2 years ago
A voltaic cell consists of a Zn&gt;Zn2+ half-cell and a Ni&gt;Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn
nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of Ni^{2+} has fallen to 0.500 M is, 0.52 V

(c) The concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

E^0_{[Ni^{2+}/Ni]}=-0.23V

E^0_{[Zn^{2+}/Zn]}=-0.76V

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

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

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

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

where,

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

E_{cell} = emf of the cell = ?

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

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

E_{cell}=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

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

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

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

E_{cell}=0.52V

(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

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

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

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0
4 years ago
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