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3 years ago

BRAINLIESTTTT ASAP!! PLEASE HELP ME :)

Chemistry

2 answers:

SIZIF [17.4K]3 years ago

8 0

Temperature is the average kinetic energy of a substance.

Murljashka [212]3 years ago

4 0

Temperature. We use it it kinetic measurements

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The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

A helium balloon is filled on the ground where the atmospheric pressure is 768 torr. The volume of the balloon is 8.00 m3. When

Anvisha [2.4K]

Answer:

The correct option is: B. 366 torr

Explanation:

Given: On the ground- Initial Volume: V₁ = 8.00 m³, Initial Atmospheric Pressure: P₁= 768 torr;

At 4200 m height- Final Volume: V₂ = 16.80 m³, Final Atmospheric Pressure: P₂ = ?

Amount of gas: n, and Temperature: T = constant

According to the Boyle's Law, for a given amount of gas at constant temperature: P₁ V₁ = P₂ V₂

⇒ P₂ = P₁ V₁ ÷ V₂

⇒ P₂ = [(768 torr) × (8.00 m³)] ÷ (16.80 m³)

⇒ P₂ = 365.71 torr ≈ 366 torr

Therefore, the final air pressure at 4200 m height: P₂ = 366 torr.

6 0

3 years ago

When a tip of a plant is cut it doesn't grow well

aleksandr82 [10.1K]

Explanation:

the tip of a growing plant contains special rapidly diving cells called apical meristem ,these cells are responsible for increase in the length of the plant . if we cut out these cells ,length growth of the plant will be stunned as these cells are not present anyplace else.

8 0

3 years ago

Read 2 more answers

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

Which of the following is characterized by existing at a discrete energy level?

hichkok12 [17]

Energy levels are considered orbitals.

5 0

3 years ago

Read 2 more answers