I'll give you three just in case a couple are ones you aren't quite looking for.
-Carbon has several allotropes, or different forms in which it can exist. A couple of allotropes included are graphite and diamond, which both have very different properties.
-Despite carbon's ability to make 4 bonds and its presence in many compounds, it is highly nonreactive under normal conditions.
-Carbon exists in three main isotopes: ¹²C, ¹³C, and ¹⁴C in which ¹⁴C is radioactive and used in dating carbon-containing samples (known as radiometric dating)
Hope this helps!!
Bk has an atomic number of 97 meaning that there are 97 protons and 152 neutrons.
When carbon is burned in air carbon iv oxide gas is formed.
C (s) + O2 (g) = CO2(g) ΔH = - 393.5 kj/mol
The enthalpy change of the reaction is -393.5 j/mol which means that when one mole of carbon is completely burnt in air then 393.5 j of energy is evolved.
Thus, 1 mole = -393.5 j , then for 480 kj
= 480 × 1/393.5
= 1.2198 moles
1 mole of carbon iv oxide is equal to 44 g
thus, 1.2198 moles will be 1.2198 × 44 = 53.6712 g of CO2
I think the answer would be d
Answer: The standard enthalpy change is -607kJ
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{C_6H_6}\times \Delta H_f^0_{(C_6H_6)}]-[n_{C_2H_2\times \Delta H_f^0_{(C_2H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BC_6H_6%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_6H_6%29%7D%5D-%5Bn_%7BC_2H_2%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_2H_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times 83)]-[(3\times 230)]=-607kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%2083%29%5D-%5B%283%5Ctimes%20230%29%5D%3D-607kJ)
The standard enthalpy change is -607kJ
