What data should be plotted to show that experimental concentration data fits a first-order reaction? g?

A 7.74 L balloon is filled with water at 3.88 atm. If the balloon is squeezed into a 0.23 L beaker and does NOT burst, what is t

Vladimir [108]

Answer:

131 atm

Explanation:

To find the new pressure, you need to use Boyle's Law:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.

P₁ = 3.88 atm P₂ = ? atm

V₁ = 7.74 L V₂ = 0.23 L

P₁V₁ = P₂V₂ <----- Boyle's Law

(3.88 atm)(7.74 L) = P₂(0.23 L) <----- Insert values

30.0312 = P₂(0.23 L) <----- Simplify left side

131 = P₂ <----- Divide both sides by 0.23

6 0

2 years ago

The first four ionization energies for an imaginary element, Xz, are E1 = 102 kcal, E2 = 186 kcal, E3 = 4021 kcal, and E4 = 4862

Anni [7]

Hi :) If your answers choices are what I'm thinking, the answer is two :)

4 0

3 years ago

Read 2 more answers

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

During the daytime, wind blows off of the __________________ because that material takes ___________ time to get warm, creating

Butoxors [25]

Answer: sea, more

I hope this helps

5 0

3 years ago

33) How many significant figures does the number 40230 have? *

Lyrx [107]

Answer:

5

Explanation:

they are all significant All non-zero numbers ARE significant

4 0

3 years ago