Data Given; T = 306 K
M.mass of CO = 28 g/mol = 0.028 Kg/mol
M.mass of CO₂ = 44 g/mol = 0.044 Kg/mol
M.mass of SO₃ = 80 g/mol = 0.080 Kg/mol
Gas Constant = R = 8.314 J/mol.K
Formula used; Vrms =
Putting values for CO; Vrms =
Vrms = 477 m.s⁻¹
Putting values for CO₂; Vrms =
Vrms = 380 m.s⁻¹
Putting values for SO₃; Vrms =
Vrms = 282 m.s⁻¹
Kinetic Energy;
K.E = 1/2 mv²
Putting values,
For CO,
K.E = 1/2 (0.028 Kg/mol) (477 m/s)²
K.E = 1/2 (0.028 Kg) (227529 m²/s²)
K.E = 3185 Kg.m².s⁻²For CO₂
,
K.E = 1/2 (0.044 Kg/mol) (380 m/s)²
K.E = 1/2 (0.044Kg) (144400 m²/s²)
K.E = 3176 Kg.m².s⁻²For SO₃
,
K.E = 1/2 (0.080 Kg/mol) (282 m/s)²
K.E = 1/2 (0.080 Kg) (79524 m²/s²)
K.E = 3180 Kg.m².s⁻²
<h2>Question:- What is the density of a cube </h2>
<h3>Given:-</h3>
- Mass = 25 grams
- Volume = 125 cm³
<h2>Answer:-</h2>
<h3>Formula :- </h3><h3>
</h3><h3>
</h3>
To convert it in Kg/m³ divide by 1000 and multiple with 10⁶
Answer:
<h3>the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Explanation:
Amount of HBr dissociated
2HBr(g) ⇆ H2(g) + Br2(g)
Initial Changes 2.15 0 0 (mol)
- 0.789 + 0.395 + 0.395 (mol)
At equilibrium 1.361 0.395 0.395 (mole)
Concentration 1.361 / 1 0.395 / 1 0.395 / 1
at equilibrium (mole/L)
<h3>Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Mass is protons + neutron so the mass number is 14
Answer:
A
Explanation:
The correct answer would be that<u> the group being given the sugar pill is being used as the control group.</u>
In a typical experimental study, there are two groups of subjects:
1. The treatment or the experimental group
2. The control group
<em>The treatment group receives the treatment variable while the control group does not receive or is given a placebo. Thus, the control group act as the reference group for comparison basis. Any difference in the observation between the two groups would be attributed to the treatment variable.</em>
The correct option is, therefore, A.