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Mamont248 [21]
3 years ago
9

5) Determine the mass if lithium hydroxide that is produced when 12.87 g of lithium

Chemistry
1 answer:
netineya [11]3 years ago
4 0

Answer: 26.54 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of lithium nitride}=\frac{12.87g}{34.83g/mol}=0.369moles

Li_3N+3H_2O\rightarrow 3LiOH+NH_3  

Li_3N is the limiting reagent as it limits the formation of product and H_2O is the excess reagent

According to stoichiometry :

As 1 moles of Li_3N give = 3 moles of LiOH

Thus 0.369 moles of O_2 give =\frac{3}{1}\times 0.369=1.108moles  of LiOH

Mass of LiOH=moles\times {\text {Molar mass}}=1.108moles\times 23.95g/mol=26.54g

Thus  26.54 g of LiOH will be produced from the given mass.

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If 3.289 x 10^23 atoms of potassium react with excess water, how many grams of hydrogen gas would be produced?
GarryVolchara [31]

Answer:

The amount in grams of hydrogen gas produced is 0.551 grams

Explanation:

The parameters given are;

Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms

Chemical equation for the reaction is given as follows;

2K + 2H₂O \rightarrow KOH + H₂

Avogadro's number, N_A, regarding the number of molecules or atom per mole is given s follows;

N_A = 6.02 × 10²³ atoms/mole

Therefore;

The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles

2 moles of potassium produces one mole of hydrogen gas, therefore;

1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas

The molar mass of hydrogen gas = 2.016 grams

Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.

The amount in grams of hydrogen gas produced = 0.551 grams.

8 0
3 years ago
What is the total number of valence electrons in a carbon atom in the ground state
kolbaska11 [484]
Carbon will have 4 valence electrons.  It will have 2 in the p orbital and 2 in the s orbital.  You can see this when you find the noble gas configuration of carbon which is [He]2s²2p² showing that carbon has 4 valence electrons.
I hope this helps.  Let me know if anything is unclear.
6 0
2 years ago
3 why does soldium produce blue colour when dissolve in ammonia?​
bogdanovich [222]
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6 0
2 years ago
Calculate the energy difference for a transition in the paschen series for a transition from the higher energy shell n=4. expres
Troyanec [42]
Electrons are orbiting around the nucleus in a specific energy level as described in Bohr's atomic model. There are 7 energy levels all in all; 1 being the strongest and nearest to the nucleus, and 7 being the weakest and farthest away from the nucleus. Electron can transfer from one energy level to another. If it increases energy, it absorbs energy. If it goes down an energy level, it emits energy in the form of light. This light can be measure in wavelength through the Rydberg equation:

1/λ =R(1/n₁² -1/n₂²), where
λ is the wavelength
R is the Rydberg constant equal to 1.097 × 10⁻7<span> per meter
n</span>₁ and n₂ are the energy levels such that n₂>n₁

In the Paschen series is an emission spectrum of hydrogen when the energy level is at least n=4. So, this covers n=4 to n=7.

1/λ =(1.097 × 10⁻7)(1/4² -1/7²)
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5 0
3 years ago
Gaseous methane ch4 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 1.44
Usimov [2.4K]
The balanced equation for the above reaction is;
CH₄ + 2O₂ ---> CO₂ + 2H₂O
Stoichiometry of CH₄ to O₂ is 1:2
The number of methane moles present - 1.44 g/ 16 g/mol = 0.090 mol
Number of oxygen moles present - 9.5 g/ 32 g/mol = 0.30 mol
If methane is the limiting reagent,
0.090 moles of methane react with 0.090x 2 = 0.180 mol 
only 0.180 mol of O₂ is required but 0.30 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.
Number of moles of water that can be produced - 0.180 mol
Therefore mass of water produced - 0.180 x 18 g/mol = 3.24 g 
Therefore mass of 3.24 g of water can be produced 
6 0
3 years ago
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