Answer:
0.02405 g/L is the solubility of argon in water at 25 °C.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant = 
= partial pressure of carbonated drink = 0.51atm
Putting values in above equation, we get:
Molar mass of argon = 39.95 g/mol
Solubility of the argon gas :
0.02405 g/L is the solubility of argon in water at 25 °C.
Weathering and erosion produce ever smaller rock particles which,
when mixed with dust and decayed organic matter over time ,
result in different types of soil .
[hope this helps]
Answer:
3 element i.e carbon (C), hydrogen (H) and oxygen.
Explanation:
The following data were obtained from the question:
Subtance >>>>>>>> Chemical Formula
Glucose >>>>>>>>> C₆H₁₂O₆
Methane >>>>>>>> CH₄
Ethanol >>>>>>>>> C₂H₅OH
Hydrogen peroxide >> H₂O₂
From the above table, we can see that ethanol (C₂H₅OH) contains carbon (C), hydrogen (H) and oxygen
Therefore, the total number of elements present in ethanol, C₂H₅OH is 3.
Answer: The correct statements are ; A and B.
Explanation:
The strength of the magnetic field(B) in an electromagnet can be calculated using formula:
= Relative permeability
N = number of turns
I = Current in the wire of the solenoid
L = length of the solenoid or electromagnetic
As we can see from the formula:
So, by increasing the turns and increasing current flowing through wire one can increase the strength of an electromagnet.
Hence, the correct statements are ; A and B.
Answer:
Elemental gold to have a Face-centered cubic structure.
Explanation:
From the information given:
Radius of gold = 144 pm
Its density = 19.32 g/cm³
Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:
a = 407 pm
In a unit cell, Volume (V) = a³
V = (407 pm)³
V = 6.74 × 10⁷ pm³
V = 6.74 × 10⁻²³ cm³
Recall that:
Net no. of an atom in an FCC unit cell = 4
Thus;
density d = 19.41 g/cm³
Similarly; For a body-centered cubic structure
where;
r = 144
a = 332.56 pm
In a unit cell, Volume V = a³
V = (332.56 pm)³
V = 3.68 × 10⁷ pm³
V 3.68 × 10⁻²³ cm³
Recall that:
Net no. of atoms in BCC cell = 2
∴
density =17.78 g/cm³
From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.
This makes the elemental gold to have a Face-centered cubic structure.
