Answer:
35.5450 will be rounded to 35.55
Explanation:
=35.5450
if the last digit is less than 5 then it will be ignored
=35.545
when the dropping digit is 5 then the retaining digit will increse by a factor of 1
=35.55
i hope this will help you
Answer is: hydrogen bonds.
Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.
According to the principle of base pairing hydrogen bonds could form between adenine and thymine (two hydrogen bonds between this nucleobases) and guanine and cytosine (three hydrogen bonds between this nucleobases).
Adenine and guanine are purine derivatives and thymine and cytosine are pyrimidine derivates.
Answer:
C₄H₁₀O + 6O₂ ⇒ 4CO₂ + 5H₂O
Explanation:
Match the amount of reactants and products on both sides of the equation.
Explanation:
A.
In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.
B.
Equation of the reaction
2NaOH + H2SO4 --> Na2SO4 + 2H2O
Number of moles of H2SO4 = molar concentration × volume
= 0.75 × 0.0105
= 0.007875 moles.
By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875
= 0.01575 moles.
Molar concentration of NaOH = number of moles ÷ volume
= 0.01575 ÷ 0.0175
= 0.9 M of NaOH.
Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
