Answer:
b. lithium
Explanation:
furthest to the left in second period
A molecular covalent substance (such as co2) has a low melting point because the covalent bonds that hold the molecules together are weak and do not require much energy to break:- False.
What are covalent bonds ?
An electron transfer that leads to the formation of electron pairs between atoms is known as a covalent bond. When atoms share electrons, a stable balance of the repulsive and attractive forces among them is known as covalent bonding. These electron pairs are also known as bonds or shared pairs.
It is a molecular compound, which is a mixture of at least two atoms—the smallest building blocks of matter—joined by a covalent bond. These atoms are joined by a covalent bond, which is formed when electrons are shared.
Learn more about covalent bond here:-
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2H2O = 2H2 + O2.
<h3><u>Explanation</u>:</h3>
Balancing equations is very essential because of the fact that it represents the stoichiometric quantities of the reactants needed to react to form the product. The ratio of the weights of reactant and product are also very well understood from this.
Here in this equation, the water is broken into hydrogen and oxygen. The balanced reaction is
2H2O = 2H2 + O2.
Two moles of water is broken down into 2 moles of hydrogen and one mole of oxygen.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
I believe this is a double replacement. As you can see none of the chemical symbols are missing, there just "replaced".
