Answer:
P = 13.5 atm
Explanation:
Given that
No. of moles, n = 20 moles
Volume of nitrogen gas = 36.2 L
Temperature = 25°C = 298 K
We need to find the pressure of the gas. Using the ideal gas equation
PV = nRT
Where
R is gas constant, 
So,
so, the pressure of the gas is equal to 13.5 atm.
1st one= CaO +H2O=Ca(OH)2
product side-
Ca=1
O=2
H=2
Reactant side-
Ca=1
O=2
H=2
The first one is balanced for you
There is 1 calcium on each side 2 oxygens on each side and 2 hydrogens on each side
Answer:
See explanation
Explanation:
An intrinsic property is a property that is internal, that is, it characterizes the substance under study. The possession of an intrinsic property depends on the nature of the substance. An intrinsic property does not depend on amount of substance but on the nature of the substance.
Examples of intrinsic properties include; Density. Solubility, Melting Point, Freezing Point, Boiling Point, Conductivity etc.
Intrinsic properties really represent the matter that is being studied. For instance, the boiling point of water will always be 100°c. No other liquid can boil exactly at that temperature. Hence, this intrinsic property can always be used to identify an unknown liquid as water.
The students were right, studying intrinsic properties accurately represent the matter that is being studied.
Answer:
D. Intramolecular covalent bond
Explanation:
Compound D is structurally more rigid as a result of intramolecular covalent bonding. The forces that hold together atoms within a compound are greater as compared to forces holding two molecules together (intermolecular bonding). On the other hand Hydrogen bonds are weaker as compared to covalent bonds. Covalent bonds involve the sharing of electrons between two atoms and Hydrogen bonds are formed between a highly electronegative atom like oxygen, Flourine,Chlorine to hydrogen.
Answer:
22.27 °C = ΔT
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass = 28 g
heat absorbed = 58 cal
specific heat of copper = 0.093 cal/g .°C
temperature change =ΔT= ?
Solution:
Q = m × c × ΔT
58 cal = 28 g × 0.093 cal /g.°C × ΔT
58 cal = 2.604 cal.°C × ΔT
58 cal / 2.604 cal .°C = ΔT
22.27 °C = ΔT
