Please Help!!

A two-liter soft drink bottle can withstand apressure of

larisa86 [58]

Explanation:

According to the ideal gas equation, PV = nRT.

where, P = pressure, V = volume

n = no. of moles, R = gas constant

T = temperature

Also, density is equal to mass divided by volume. And, no. of moles equals mass divided by molar mass.

Therefore, then formula for ideal gas could also be as follows.

P = $\frac{mass}{volume \times molar mass} \times RT$

or, P = $\frac{density}{\text{molar mass}} \times RT$

Since, density is given as 0.789 g/ml which is also equal to 789 g/L (as 1000 mL = 1 L). Hence, putting the given values into the above formula as follows.

P = $\frac{density}{\text{molar mass}} \times RT$

= $\frac{789 g/l}{46.06 g/mol} \times 0.0821 L atm/mol K \times 373 K$

= 525 atm

As two-liter soft drink bottle can withstand a pressure of 5 atm and the value of calculated pressure is 525 atm which is much greater than 5 atm.

Therefore, the soft drink bottle will obviously explode.

4 0

3 years ago

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42×10−4s−1 at a certain temperature. What is t

Keith_Richards [23]

Answer:

a) Half life of the decomposition = 4951.1 s ≈ 4950 s

b) Time it will take for the concentration of SO₂Cl₂ to decrease to 25% of its initial concentration = 9900 s

c) If the initial concentration of SO₂Cl₂ is 1.00 M, time it will take for the concentration to decrease to 0.78 M is 1775s

d) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.146 M

e) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.1398 M

Explanation:

Let C₀ represent the initial concentration of SO₂Cl₂

And C be the concentration of SO₂Cl₂ at anytime.

a) Rate of a first order reaction is represented by

dC/dt = - KC

dC/C = - kdt

Integrating the left hand side from C₀ to C₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)

In [(C₀/2)/C₀] = - k t(1/2)

In (1/2) = - k t(1/2)

- In 2 = - k t(1/2)

t₍₁,₂₎ = (In 2)/k

t₍₁,₂₎ = (In 2)/(1.4 × 10⁻⁴)

t₍₁,₂₎ = 4951.1 s

b) dC/C = - kdt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = - kt

C/C₀ = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C = 25% of C₀ = 0.25C₀

0.25C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

- kt = In 0.25

- kt = - 1.386

t = 1.386/(1.4 × 10⁻⁴) = 9900 s

c) C = C₀ e⁻ᵏᵗ

C = 0.78 M; C₀ = 1.00 M

0.78 = 1 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.78

- kt = In 0.78

- kt = - 0.2485

t = 0.2485/(1.4 × 10⁻⁴) = 1775 s

d) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 2 × 10² s = 200 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 200) = 0.972

C = 0.15 × 0.972 = 0.146 M

e) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 5 × 10² s = 500 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 500) = 0.9324

C = 0.15 × 0.9324 = 0.1398 M

6 0

3 years ago

Help quick! Will mark brainlest!

Ivenika [448]

Answer: 10HSiCl3 + 15H2O → H10Si10O15 + 30HCl

Explanation:

6 0

2 years ago

What is 2(NH,4)Cr,2 O,7

Vitek1552 [10]

oxidation-reduction.

4 0

3 years ago

Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42

Neko [114]

The pH of the equivalence points is 8.54

Concept of pH

A solution's acidity or alkalinity can be determined based on the concentration of hydrogen ions in the solution, or pH. Acidic aqueous solutions at 25 °C have a pH under 7, while basic or alkaline aqueous solutions have a pH above 7. Since the concentration of H3O+ is equal to the concentration of OH in pure water, a pH level of 7.0 at 25°C is referred to as "neutral". Strong bases may have a pH above 14, while very strong acids may have a pH below 14.

0.0520 M CH3COOH in 42.2 mL of moles is as follows:

2.194x103 mol CH3COOH = 0.0422L (0.0520mol / L)

that react with NaOH, resulting in:

NaOH + CH3COOH = CH3COO + Na+ + H2O

Thus, 1 mole of acetic acid and 1 mole of NaOH react.

As a result, 2.194x103 mol of NaOH are required to reach the equivalence point in volume:

To attain the equivalency point, 2.194x103 mol (1L / 0.0372mol) = 0.05899L 58.99mL of 0.0372 M NaOH

You will only have CH3COO at the equivalency point because it is in equilibrium with water, so:

H2O(l) + CH3COO(aq) CH3COOH(aq) + OH (aq)

A definition of equilibrium is:

Kb = 5.6x1010 = [OH] / [CH3COO] / [CH3COOH]

2.194x103mol of CH3COO has a molarity of (0.05899L + 0.0422L) = 0.02168M.

Therefore, equilibrium concentrations are:

[CH3COO]=0.02168M-X [CH3COOH]=X [OH]=X

5.6x1010 = [X] [X] / [0.02168M - X] converts to Kb.

1.214x1011 - 5.6x1010X = X2 X2 + 5.6x1010X - 1.214x1011 = 0 Finding the value of X:

False response; there are no negative concentrations. X: -3.48x106

As [OH] = X, [OH] = 3.484x106M, X is 3.484x106.

As 14 = pOH + pH pH = 8.54, pOH = -log [OH], or 5.46.

To know more about pH visit :

brainly.com/question/12546875

#SP4

3 0

2 years ago