Solids maintain their shape, whereas fluids do not because the molecules in solids maintain a regular pattern and only vibrate, or move very slowly
Answer:
H₂O + CO₂ → H₂CO₃
Option D is correct.
Law of conservation of mass:
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two H and two O atoms present on left side while on right side only one O and two H atoms are present so mass in not conserved. This option is incorrect.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side of equation while on right side two H, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. This option is correct.
Answer:
HUH I NEED A PIC of the question
Explanation:
The volume of a gas that its pressure increase to 3.4 atm is calculated as follows
By use of boyles law that is P1V1=P2V2
V1=4.0 L
P1=1.1 atm
P2=3.4 atm
V2= P1V1/P2
(1.1 atm x 4.0 L)/3.4 atm= 1.29 L
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
