First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
Answer:
CH₄
Explanation:
To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.
Step 1: Determine the mass of the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g
Step 2: Calculate the percent by mass of each element
%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%
%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%
Step 3: Divide each percentage by the atomic mass of the element
C: 74.8/12.01 = 6.23
H: 25.2/1.01 = 24.95
Step 4: Divide both numbers by the smallest one, i.e. 6.23
C: 6.23/6.23 = 1
H: 24.95/6.23 ≈ 4
The empirical formula of the hydrocarbon is CH₄.
Answer:
The answer is 37amu
Explanation:
We are given:
Chlorine-35: atomic mass =35 amu and percent abundance =75.5% ≡ 0.755
Chlorine-37: atomic mass = x amu and percent abundance =24.5% ≡ 0.25
From table, relative atomic mass of Chlorine is 35.5
⇒(0.755 × 35) + (24.5 × x) = 35.5
26.402 + 24.5x = 35.5
∴ x = 9.075 ÷ 24.5 = 0.37 ≡ 37.
∴ mass of Chlorine 37 = 37amu
None of these; the most electronegative element is fluorine because an atom of fluorine needs an additional electron to join it's outer membrane, so the atom can achieve stability.
a flourine atom on its own is always a negative ion.
