CH₇ is the empirical formula of the car fuel.
Explanation:
To find the empirical formula we use the following algorithm.
First divide each mass the the molar weight of each element:
for carbon 2.87 / 12 = 0.239
for hydrogen 3.41 / 2 = 1.705
And now divide each quantity by the lowest number which is 0.239:
for carbon 0.239 / 0.239 = 1
for hydrogen 1.705 / 0.239 = 7.13 ≈ 7
The empirical formula of the car fuel is CH₇.
I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.
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PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72
I hope this helps. Let me know if anything is unclear.
Number of moles= mass/ molar mass
Or n=m/MM
n = number of moles
m = mass
MM = molar mass
1) n CuO = 2.4g / 79.54g/mol = 0.03 mol CuO
2) n Cu(NO3)2.xH2O = 7.26 g / 205.6 = 0.035 moles of Cu(NO3)2.xH2O
3) 205.6 g
Cu = 63.5 g
N = 14g
O =16g
H= 1 g
63.5+ (14+(16*3))*2+1*2+16 =205.6 g
4) yes is 188g
5) I don’t know, I assume was 1
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) =
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
The four ionic species initially in solution are Na⁺, PO₄³⁻, Cr³⁺, and Cl⁻. Since the precipitate is composed of Cr³⁺ and PO₄³⁻ ions, the spectator ions must be Na⁺ and Cl⁻.
The complete ionic equation is 3Na⁺(aq) + PO₄³⁻(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → 3Na⁺(aq) + 3Cl⁻(aq) + CrPO₄(s).
So the balanced <u>net ionic equation</u> for this reaction would be Cr³⁺(aq) + PO₄³⁻(aq) → CrPO₄(s).