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2 years ago

Determine the missing species:

Chemistry

1 answer:

Galina-37 [17]2 years ago

5 0

Answer: 4. $^{0}_{-1}\textrm{e}$ and $^{0}_{1}\textrm{e}$

Explanation:

a) The given reaction is $^{41}_{20}\textrm{Ca}+^{x}_{y}\textrm{X}\rightarrow ^{41}_{19}\textrm {K}$

As the mass on both reactant and product side must be equal:

$41+x=41$

$x=0$

As the atomic number on both reactant and product side must be equal:

$20+y=19$

$y=-1$

$^{41}_{20}\textrm{Ca}+^{0}_{-1}\textrm{e}\rightarrow ^{41}_{19}\textrm {K}$

b) $^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{x}_{y}\textrm{X}$

Total mass on reactant side = total mass on product side

15 =15 + x

x = 0

Total atomic number on reactant side = total atomic number on product side

8 = 7 + y

y = 1

$^{15}_{8}\textrm{O}\rightarrow ^{15}_{7}\textrm{N}+^{0}_{1}\textrm{e}$

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Classify the following aqueous solutions as: strong acid, weak acid, neutral, weak base, or strong base.

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Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take

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Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

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Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = 165°C

Depression in freezing point = $\Delta T_f=?$

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Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

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m = molality of the sample

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i = 1 ( organic compounds)

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3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

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