Answer:
Explanation:
Diamond has lesser density than platinum . So , if we take equal mass of both , the volume of mass of platinum will be far less .
The density of both diamond and platinum are more than water so both of them will be drowned in water completely . They will not float . On being drowned , platinum will displace lesser volume of water because of its less volume . So volume change in case of platinum mass will be far less . The volume change for diamond will be more because of its bigger size.
The mass of oxygen reacted/required in this reaction is obtained as 48g.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.
Now, we have the reaction; S + O2 ------>SO2
If 1 mole of sulfur dioxide contains 22.4 L
x moles of sulfur dioxide contains 33.6L
x = 1.5 moles of sulfur dioxide.
Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.
Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g
Learn more anout stoichiometry: brainly.com/question/9743981
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:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
Answer:
Please see the attached pictures.
Explanation:
☆ To ensure that each carbon has 4 bonds, fill the other bonds with Hs.