Answer : The mole ratios of Hydrazine to Hydrogen peroxide is 1 : 2 and
the mole ratios of Hydrazine to water is 1 : 4.
Explanation :
The balanced chemical equation is,
According to the given reaction,
1 mole of Hydrazine react with the 2 moles of Hydrogen peroxide.
Therefore, the mole ratios of Hydrazine to Hydrogen peroxide is 1 : 2
And in case of Hydrazine and water,
1 mole of Hydrazine gives 4 moles of water.
Therefore, the mole ratios of Hydrazine to water is 1 : 4
Answer:
It is 20. g HF
Explanation:
H2 + F2 ==> 2HF ... balanced equation
Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).
moles of H2 present (using Avogadro's number):
3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2
From the balanced equation, we see that 1 mole H2 produces 2 moles HF. Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:
0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.
The molar mass of HF = 20.01 g/mole, thus...
0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)
Lewis diagram:
well first you have to know the Valence electrons the elements which will on top of the periodic table.
for examples:
H2O
it will be O-H-O and the dots will on top. oxygen have 6 valence electrons so that mean it need 6 dots. also for H it will have 7 V-electrons.
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A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is (
).
Utilize the titration method of in view that we're given the concentrations of every compound and the quantity of 
. Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.
- M1 = initial mass
- V1= initial volume
- M2 = final mass
- V2= final volume
- (0.138)(V1) = (0.205)x(26.0)
- V2=(0.205)x(26.0)\ 0.138
- V2 = 47.10 M/L
- The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L
Read more about the neutralization:
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