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cestrela7 [59]
2 years ago
8

Iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii jdjdj

Chemistry
2 answers:
Bogdan [553]2 years ago
4 0

Answer:

lol thanks for the points <33

Feliz [49]2 years ago
3 0
Hi! How is your day?
You might be interested in
List earth and the gas giant planets from the hottest to the coldest planet
inessss [21]

Answer:

1. Venus

471°C

2. Mercury

(430°C) during the day,  (-180°C) at night

3. Earth

16°C

4. Mars

-28°C

5. Jupiter

-108°C

6. Saturn

-138°C

7. Uranus

-195°C

8. Neptune

-201°C

Explanation:

.

8 0
1 year ago
What is the volume of 0.1 mole of methane (CH4) ? (One mole of any gas occupies 22.4 L under certain conditions of temperature a
Veronika [31]

Answer:

Option A = 2.2 L

Explanation:

Given data:

volume of one mole of gas = 22.4 L

Volume of 0.1 mole of gas at same condition = ?

Solution:

It is known that one mole of gas at STP occupy 22.4 L volume. The standard temperature is 273.15 K and standard pressure is 1 atm.

For 0.1 mole of methane.

0.1/1 × 22.4 = 2.24 L

0.1 mole of methane occupy 2.24 L volume.

8 0
3 years ago
Read 2 more answers
List a possible set of four quantum numbers (n,l,ml,ms) in order, for the highest energy electron in gallium?
Elena-2011 [213]
Gallum: Z = 31

electron configuration: [Ar] 4s^2 3d10 4s2 4p1

Highest energy electron: 4p1

Quantum numbers:

n = 4, because it is the shell number
l = 1, it corresponds to type p orbital 
ml = may be -1, or 0, or +1, depending on space orientation, they correspond to px, py, pz
ms = may be -1/2 or +1/2, this is the spin number. 
7 0
2 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
A chemistry student was asked to calculate the number of moles of iron required to react with 1.20 mol of oxygen to produce iron
Delicious77 [7]

Answer:Write and balance the equation

4Fe + 3O2 -> 2Fe2O3

0.32 mol Fe x 2 mol Fe2O3 / 4 mol Fe =

0.16 mol of Fe2O3

Explanation:

8 0
2 years ago
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