Covalent network. <span>A solid that is extremely hard, that has a very high melting point, and that will not conduct electricity either as a solid or when molten is held together by a continuous three-dimensional network of covalent bonds. Examples include diamond, quartz (SiO </span><span>2 </span>), and silicon carbide (SiC). The electrons are constrained in pairs to a region on a line between the centers of pairs of atoms.<span>
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Answer:
The correct answer is - yes, 4.57 g of solute per 100 ml of solution
Explanation:
The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.
Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.
Then solublity will be :
32. g solute / 700 ml solution = y / 100 ml solution
⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.
Thus, the answer is 4.57 g of solute per 100 ml of solution.
Answer:
Covalent solids, also called network solids, are solids that are held together by covalent bonds. As such, they need localized electrons (shared between the atoms) and therefore the atoms are arranged in fixed geometries. Distortion far from this geometry can only occur through a breaking of covalent sigma bonds.
Pls help i need one more brainly to rank up. And have a great day :D
<span>i think its Uranium Dating </span>
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
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2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
