To connect the parts of a circuit and transport electricity in it from one place to nother
To do this problem, we must first look at the balanced chemical equation for the decomposition of potassium chlorate:
<span>2KClO3 --> 2KCl + 3O2 </span>
<span>We can take the given amount of grams, and use the molar mass of KClO3 to convert to moles. Then, we can use the stoichiometric ratios to relate moles of KClO3 to moles of O2. </span>
<span>(39.09)+(35.45)+(3*15.99)= 122.51 g/ mol = molar mass of KClO3 </span>
<span>45.8 g KClO3/ 122.51 g/ mol KClO3 = .374 moles KClO3 </span>
<span>.374 mol KClO3 *(3 moles O2/2 mol KClO3)= .560 moles O2 </span>
<span>Once we have moles of O2, we can convert to grams of O2. </span>
<span>(2*15.99)= 31.98 g/mol = molar mass of O2 </span>
<span>(.560 moles O2) (31.98 g/mol)= 17.91 g O2 </span>
<span>Hope this helps :)</span>
Metals:
<span>Distinguishing luster (shine)
</span><span>Malleable and ductile (flexible) as solids
</span><span>Conduct heat and electricity
</span><span>Metallic oxides are basic, ionic
</span><span>Cations in aqueous solution
</span>
Nonmetals:
<span>Non-lustrous, various colors
</span><span>Brittle, hard or soft
</span><span>Poor conductors
</span><span>Nonmetallic oxides are acidic, compounds
</span><span>Anions, oxyanions in aqueous solution
</span>
Answer:
1) Heterogeneous mixture
2). homogeneous mixture
3) water is solvent and sugar is solute
4). sublimation process
Answer :
a) The concentration of 
at equilibrium is 0.08 m.
b) The concentration of 
at equilibrium is 0.26 m.
Solution :
The equilibrium reaction is,
![Fe^{3+}+SCN^-\rightleftharpoons [FeSCN]^{2+}](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%2BSCN%5E-%5Crightleftharpoons%20%5BFeSCN%5D%5E%7B2%2B%7D)
Initial concentration 0.230 0.410 0
At equilibrium (0.230 - x) (0.410 - x) x
Given x = 0.150
Therefore,
The concentration of = 0.230 - x = 0.230 - 0.150 = 0.08
The concentration of = 0.410 - x = 0.410 - 0.150 = 0.26
Thus, the concentration of at equilibrium is 0.08 m.
The concentration of at equilibrium is 0.26 m.
