Answer:
The bond between XY is Covalent
Explanation:
Types of Bonds can be predicted by either identifying the two bonding elements or by calculating the difference in electronegativity.
The two atoms bonded to each other can be both metals, non-metals or one metal and the other non-metal. If both are metals then we can say that the bond is metallic. If both are non-metals then we can say that the bond is nonpolar and covalent. If one is metal and the other is non-metal then the bond between them will be ionic.
Secondly,
If, Electronegativity difference is,
Less than 0.4 then it is Non Polar Covalent Bonding
Between 0.4 and 1.7 then it is Polar Covalent Bonding
Greater than 1.7 then it is Ionic Bonding
As, the element X and Y have the electronegativity difference of 1.2 (which is less than 1.7 and greater than 0.4) therefore, we can say that the bond between them is covalent and it is polar in nature.
Answer:
Visible light contains all the colors from violet to red. An object gets its color when electrons absorb energy from the light and become “excited” (raised to a state of increased energy). The excited electrons absorb certain wavelengths of light.
The Sun emits light of every possible frequency at once, including at frequencies too high or too low for us to see. But the Sun's highest intensity radiation aligns approximately with our visible range – red through blue. That's no coincidence – like all animals on Earth, we have evolved to make best use of the light available.
Explanation:
Answer:
nonmetal
________ is an atom that gains electrons; it's valence electrons is similar to a noble gas.
Explanation:
Because non metal like chlorine gain one elctron and attain electronic configuration of noble gas Argon
Answer:
ΔrxnH = -580.5 kJ
Explanation:
To solve this question we are going to help ourselves with Hess´s law.
Basically the strategy here is to work in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.
Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-
The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.
This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.
Taking 1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.
Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l) ΔrxnHº = 259 / 2 kJ 1/2 eq3
+ 2NaOH(s) + SO₃(g) ⇒ Na₂SO₄ (s) + H₂O (l) ΔrxnHº = -418 kJ - eq 2
+ 2Na (s) + 2 H₂O (l) ⇒ 2 NaOH (s) + H₂ (g) ΔrxnHº = -146 x 2 2 eq 1
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Na₂O (s) + SO₃ (g) ⇒ Na₂SO₄ (s) ΔrxnHº = 259/2 + (-418) + (-146) x 2 kJ
ΔrxnH = -580.5 kJ