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3 years ago

What is the boiling point of 5 ml of alcohol

Chemistry

1 answer:

NeTakaya3 years ago

4 0

173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it

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Calculate the mass of an object that has a density of 3.52g/cm^3 and

valkas [14]

Answer:

17.6kg

Explanation:

mass = Volume × Density

mass = 5000cm^3 × 3.52g/cm^3

mass = 17600g = 17.6kg

3 0

4 years ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

4 years ago

The pressure in car tires is often measured in pounds per square inch (lb/in.2), with the recommended pressure being in the rang

Minchanka [31]

Answer:

2.62 atm

Explanation:

1 atm = 14.6959 psi

38.5 / 14.6959 = 2.62 atm

4 0

2 years ago

Barium has a body-centered cubic structure. If the atomic radius of barium is 222 pm, calculate the density of solid barium?

Flauer [41]

Answer: density =3.377g/cm³

Explanation:

Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)

D= (M ×n) /(V×A)

M= 137g/mol

n= 2 (For BCC)

V=a³ , where a= 4r/√3

a= (4×222)/√3

a=512.69pm

a= 512.69×10^-10cm

V= ( 512.69×10^-10)^3

V= 1.3476×10^-22cm³

D= (137×2)/(1.3476×10^-22 × 6.02^23)

D= 3.377g/cm³

Therefore the density of barium is 3.377g/cm³

5 0

3 years ago

A sample of o2 gas (2.0 mmol) effused through a pinhole in 5.0 s. it will take __________ s for the same amount of co2 to effuse

s344n2d4d5 [400]

Answer is: 5,9.
Effusion is leakage of gas through a small hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. Relative rates of effusion is related to the molecular mass.
M(O₂) = 32g/mol.
M(CO₂) = 44g/mol.
t₁ = 5s.
t₂ = ?
t₂/t₁ = √(M(CO₂)/M(O₂)).
t₂ = 1,17·5s = 5,9s.

8 0

4 years ago

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