<h3>
Answer:</h3>
78.34 g
<h3>
Explanation:</h3>
From the question we are given;
Moles of Nitrogen gas as 2.3 moles
we are required to calculate the mass of NH₃ that may be reproduced.
<h3>Step 1: Writing the balanced equation for the reaction </h3>
The Balanced equation for the reaction is;
N₂(g) + 3H₂(g) → 2NH₃(g)
<h3>Step 2: Calculating the number of moles of NH₃</h3>
From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃
Therefore, the mole ratio of N₂ to NH₃ is 1 : 2
Thus, Moles of NH₃ = Moles of N₂ × 2
= 2.3 moles × 2
= 4.6 moles
<h3>Step 3: Calculating the mass of ammonia produced </h3>
Mass = Moles × molar mass
Molar mass of ammonia gas = 17.031 g/mol
Therefore;
Mass = 4.6 moles × 17.031 g/mol
= 78.3426 g
= 78.34 g
Thus, the mass of NH₃ produced is 78.34 g
Carbohydrates. If you think about it, it's a mix of the three words.
The ML of 0.85 m NaOH required to titrate 25 ml of 0.72m hbr to the equivalence point is calculated as follows
calculate the moles of HBr used
moles = molarity x volume
25 x0.072/1000= 0.0018 moles
write the equation for reaction
NaOH + HBr = NaBr + H2O
from reacting equation the mole ratio between NaOH to HBr is 1:1 therefore the moles of NaOH = 0.0018 moles
volume = moles/molarity
0.0018/0.085 = 0.021 L in Ml = 0.021 x1000=21.18 Ml ofNaOH
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
Answer:
The pH decreases.
Explanation:
Hello,
In organic chemistry, oxidation accounts for either the increasing of C-O bonds or the increasing in the oxygen atoms into the molecule. Thus, if we consider the oxidation from benzyl alcohol to benzoic acid, there will be a carboxyl functional group instead of a hydroxile one. Now, the presence of the polar -O-H bonds that are ionizable, there will be a 
releasing causing the pH to decrease (increase acidity).
Regards.
