Answer:
the answer is 1s22s22p63s23p1
It is representing syncline rock formation
there are two rock formation based on fold formation : syncline and anticline
In syncline rock formation there is fold like trough unlike anticline where it is like crust
In syncline the fold is downward as shown in photo and the new rock is outer fold and old at inner side
1/2=5750 years, 1/2(1/2)=1/4, (1/2)(1/2)(1/2)=1/8, (1/2)(1/2)(1/2)(1/2)=1/6
4 halflives have passed so 4(5750)=23000 years since the tree was chopped down
1000000 atoms (1/2)=500000 atoms(1/2)=250000(1/2)=125000(1/2)=62500 atoms would remain in the wood after 4 halflives
Dinosaurs became extinct around 62 million years ago, so if 14C's half life has a value of 5750 years, it would be gone or in such small amounts that dating would be ineffective today.
As Potassium decays into Argon in 1.3 billion years, apart from volcanic activity, it would enable geologists to effectively date things that are really, really, really old.
Answer:
A)
Cl + O3 --> ClO + O2
2ClO --> ClOOCl
ClOOCl --> 2 Cl + O2
B)
Cl2 + 4O3 -> 2 ClO + 5O2
Explanation:
Step 1: Chlorine atoms react with ozone (O3) to form chlorine monoxide and molecular oxygen
Cl + O3 --> ClO + O2
Step 2: Chlorine monoxide molecules combine to form ClOOCl gas
2ClO --> ClOOCl
Step 3: ClOOCl absorbs sunlight and breaks into chlorine atoms and molecular oxygen
ClOOCl --> 2 Cl + O2
B) Overall Balanced equation
Cl2 + 4O3 -> 2 ClO + 5O2
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
<em>Where q is heat of reaction in J,</em>
<em>m is the mass of the solution in g,</em>
<em>S is specific heat of the solution = 4.184J/g°C</em>
<em>ΔT is change in temperature = 11.21°C</em>
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
