Many insulating materials contain spaces filled with air because ; Air is a poor conductor of heat and electricity
Insulators are materials that do not allow the easy passage of heat or electricity through them. they are used to protect users from direct exposure to electricity and heat and also for the protection of equipment's at home and at work places.
Insulators are generally bad conductors of heat and electricity. some examples of insulators are ;
They are all very good insulators. but the most commonly used insulator in electrical installations is Porcelain
Hence we can conclude that many insulating materials contain spaces filled with air because Air is a poor conductor of heat and electricity.
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Answer:
lithium chloride
Explanation:
Li=Lithium
Cl=Chlorine
The correct name for LiCl is Lithium Chloride. It's an ionic compound composed of Lithium and Chloride ions, arranged in a lattice structure with Chloride ions contacting each other across diagonally on the face of a unit cell.
The correct name for LiCl is simply Lithium Chloride. This is an ionic compound, which means it is composed of two ions: the lithium ion (Li+) and the chloride ion (Cl-). In this case, Li (Lithium) loses an electron and becomes a cation (Li+), whereas Cl (Chlorine) gains that electron and becomes an anion (Cl-). The compound LiCl is then formed from the electrostatic attraction between these ions.
On the face of a LiCl unit cell, Chloride ions contact each other across the diagonal of the face, showing the close packing arrangement of the ions in the lattice structure of the compound.
It's worth noting that the common naming system for these types of compounds is quite straightforward. The name of the positive ion (the cation) comes first (Lithium), followed by the name of the negative ion (the anion) with an -ide suffix (Chloride).
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Answer:
16.4 L
Explanation:
we can use the combined gas law equation that gives the relationship among volume, temperature and pressure conditions of gases.
P1V1/T1 = P2V2/T2
STP conditions are standard temperature and pressure conditions
P1 is standard pressure = 1 atm , T1 is standard temperature = 273 K
and V1 is the volume
P2 is pressure, T2 is temperature and V2 is volume at the second instance
temperature is in kelvin scale,
512 ° + 273 = 785 K
substituting the values in the equation
1 atm x 10.0 L / 273 K = 1.75 atm x V / 785 K
V = 16.4 L
new volume is 16.4 L
what is the reacted volume of oxygen gas at STP and the mass of the produced water vapour in this reaction ?
Answer:
Volume of O₂ = 56 dm³
mass of water vapors (H₂O) = 90 g
Explanation:
Data Given:
mass of Oxygen = 10 g
Volume of Oxygen = ?
mass of the water vapor = ?
Reaction Given:
2H₂+O₂---->2H₂O
Solution:
First we have to look at the reaction for the information required
2H₂ + O₂ -------> 2H₂O
2 mol 1mol 2 mol
now convert moles to grams
molar mass of H₂ = 2(1) = 2 g/mol
molar mass of O₂ = 2(16) = 32 g/mol
molar mass of H₂0 = 2(1) + 16 = 18 g/mol
So the masses will be
2H₂ + O₂ -------> 2H₂O
2 mol (2 g/mol) 1mol (32 g/mol) 2 mol (18 g/mol)
4 g 32 g 36 g
So now we know that
4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.
By using above information
First we find the volume of Oxygen:
For this first we find mass and then moles of Oxygen
As we know
if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen
Apply unity formula
4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂
10 g of hydrogen H₂ ≅ X g of Oxygen O₂
by doing Cross multiplication
g of Oxygen O₂ = 32 g x 10 g / 4 g
g of Oxygen O₂ = 80 g
So,
mass of oxygen = 80 g
now find moles of oxygen
formula used:
no. of moles = mass in grams/ molar mass . . . . . . (1)
Put values in above equation 1
no. of moles = 80 g / 32 g/mol
no. of moles = 2.5
Now to find volume of oxygen
Formula used
Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)
Put values in equation 2
Volume of O₂ = 2.5 moles x 22.4 dm³/mol
Volume of O₂ = 56 dm³
______________________
Now to find mass of water vapors
As we now
if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen
Apply unity formula
4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)
10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)
by doing Cross multiplication
g of water vapors (H₂O) = 36 g x 10 g / 4 g
g of water vapors (H₂O) = 90 g
So,
mass of water vapors (H₂O) = 90 g