Please help me it due today at 11:00am please help me will mark the brainiest please

Chemistry

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

Point f because that is when it starts going down

You might be interested in

Which statement correctly describes two forms of oxygen, O2 and O3?

fgiga [73]

The two forms of oxygen, O2 and O3 is "<span>They have different molecular structures and different properties."</span>

7 0

3 years ago

Read 2 more answers

Density,mass,volume: Is the amount of matter a substance has

LuckyWell [14K]

Density is the mass of a substance per unit volume. Volume is the amount of space an object occupies. Chemical properties- These are properties that can only be observed by changing the identity of the substance

Hope it helped ya out!

5 0

4 years ago

The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K

BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + $\int\limits^{T2}_{T1} {DCp(T)} \, dT$

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

$\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT$ = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

7 0

3 years ago

At 50.14k a substance has a vapor pressure of 258.9 torr. calculate its heat of vaporization in kj/mol if it has a vapor pressur

gayaneshka [121]

T₁ = 50,14 K.
p₁ = 258,9 torr.
T₂ = 161,2 K.
p₂ = 277,5 torr.
R = 8,314 J/K·mol.
Using Clausius-Clapeyron equation:
ln(p₁/p₂) = - ΔHvap/R · (1/T₁ - 1/T₂).
ln(258,9 torr/277,5 torr) = -ΔHvap/8,314 J/K·mol · (1/50,14 K - 1/161,2 K).
-0,069 = -ΔHvap/8,314 J/K·mol · (0,0199 1/K - 0,0062 1/K).
0,0137·ΔHvap = 0,573 J/mol.
ΔHvap = 41,82 J.

5 0

4 years ago

: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be

Jet001 [13]

Answer: New concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.