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3 years ago

A ____usually requires calculations that are too complicated to do byhand.

Chemistry

1 answer:

liraira [26]3 years ago

8 0

Answer:

computer model

Explanation:

Computer models are cheaper to set up than alternative methods that could be used to predict what will happen in a system, ex. building a prototype. Other benefits include being able to: make alterations and quickly see the outcomes.

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A. True <br> b. False a chemical species is either an acid or a base, it cannot function as both

ivann1987 [24]

False They can function as both. An example is Aluminium Oxide. These kind of substances are called "Amphoteric", they can behave as both acids and bases.

3 0

3 years ago

Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

3 years ago

Consider a certain type of nucleus that has a half-life of 32 min. calculate the percent of original sample of nuclides remainin

Irina18 [472]

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.

32 = 0.693 / λ
λ = 0.693 / 32 (1)

Nt = Nο eΛ(-λt) (2)

Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min

From (1) and (2),

Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt = 0.085Nο

Percentage = (Nt/Nο) x 100%
= (0.085Nο/Nο) x 100%
= 8.5%

Hence, Percentage of remaining atoms with the original sample is 8.5%

8 0

3 years ago

What are two characteristic properties of chlorine that distinguish it from oxygen?

sesenic [268]

Well one characteristic gases and the state of matter(one of the distinct form i which matter exist)

3 0

3 years ago

For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con

Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation: