Answer:
The answer to your question is below
Explanation:
In sodium chloride, its atoms are attached by ionic bonding.
In sucrose, its atoms are attached by covalent bonding.
Characteristics of substances that have ionic bonding high melting and boiling points, they conduct electricity in solution, are crystals, ionic bonding is stronger than covalent bonding.
Characteristics of substances that have covalent bonding have low melting and boiling points, they do not conduct heat or electricity, covalent bonding is a weak bond.
According to the previous description, the correct answer is " Sodium chloride has a higher boiling point than sucrose"
Answer: C-N(Longest) > C=N > CN(C triple bond N, shortest)
Explanation: Bond length is the distance between nuclei of bonded atoms.
Bond energy is the which needs to break the bond.
And bond length is always inversily proportional to bond energy.
Larger the bond energy ,shorter the bond length
B: SO2
S=Sulfur
O=Oxygen
Dioxide if i am correct means 2 oxygen.
so thats one Sulfur and 2 oxygen.
With that informatio you can:
1) Write the chemical equation
2) Balance the chemical equation
3) State the molar ratios
4) Predict if precipitation occurs.
I will do all four, for you:
1) Chemical equation:
mercury(I) nitrate potassium bromide mercury(I) bromide potassium nitrate
<span>Hg2(NO3)2 + KBr → Hg2Br2 + KNO<span>3
2) Balanced chemical equation
</span></span>
<span>Hg2(NO3)2 + 2KBr → Hg2Br2 + 2KNO<span>3
3) Molar ratios or proportions:
1 mol </span></span><span>Hg2(NO3)2 : 2 mol KBr : 1 mol Hg2Br2 : 2 mol KNO<span>3
4) Prediction of precipitation.
You can use the solubility rules or a table of solubilities. I found in a table of solutiblities that mercury(I) bromide is insoluble and potassium bromide is soluble, Then you can predict that the precipitation of mercury(I) bromide will occur.
</span></span>
Answer:
Energy of photon is 34.45 ×10⁻¹⁹ J.
Explanation:
Given data:
Frequency of photon = 5.2×10¹⁵ s⁻¹
Plancks constant = 6.626×10⁻³⁴ Js
Energy of photon = ?
Solution:
Formula:
E = h.f
by putting values,
E = 6.626×10⁻³⁴ Js × 5.2×10¹⁵ s⁻¹
E = 34.45 ×10⁻¹⁹ J
Thus, energy of photon is 34.45 ×10⁻¹⁹ J.
