Please help and thank you

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

I need help can someone please do so?

larisa86 [58]

Answer:

0.296 J/g°C

Explanation:

Step 1:

Data obtained from the question.

Mass (M) =35g

Heat Absorbed (Q) = 1606 J

Initial temperature (T1) = 10°C

Final temperature (T2) = 165°C

Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C

Specific heat capacity (C) =..?

Step 2:

Determination of the specific heat capacity of iron.

Q = MCΔT

C = Q/MΔT

C = 1606 / (35 x 155)

C = 0.296 J/g°C

Therefore, the specific heat capacity of iron is 0.296 J/g°C

8 0

3 years ago

Many appliances in your home and school use electrical energy. Which of these appliances is intended to convert electrical energ

Solnce55 [7]

Answer:

music amplifier or a loudspeaker

Explanation:

A music amplifier or a loudspeaker are a few devices that convert electrical energy into sound energy.

As it is a well known fact that energy can neither be created nor be destroyed it can only change from one form to the other. As here it is changing from electrical to sound.

6 0

2 years ago