Answer:
True
Explanation:
when a physical change occurs in a substance, the appearance might change but the chemical identity remains same. For example, if liquid water is heated, the water changes to gaseous form .This means the appearance of the water has changed, but the chemical identity remains same as both liquid water and steam has both oxygen and hydrogen as the component molecules or atoms.
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>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 <em>law of ideal gases,</em>
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
x
→ 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 = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>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
x
x 100%
c = 0.003 %
So, <u>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 </u>27.09 ppm and 0.003 %.
Answer: 2.52*10^25
Explanation:
6.02214076*10^23 in 1 mole
Answer:
Repetition
Explanation:
The best term to describe this example is repetition and not replication.
Most times, to obtain an accurate and precise reading which is reliable, scientist makes several repeated measurements. The average gives the most reliable representation of the phenomenon being tested.
This process is called repetition.
On the other hand, duplicating an experiment is replication. Experiments are replicated to proof their validity.
- Since this is a brand new experiment where 16 height measurements are reported, we are dealing with a repetition situation.
Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:


We make a simplification because x<<< 0.0225:

x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M