The formula of the ppt. formed is PbSo4 , which is inslouble.
Answer:
The answer to your question is: letter B
Explanation:
Data
3.58 x 10⁴ cal to joules
1 calories --------------- 4.184 joules
3.58 x 10⁴ cal----------- x
x = ( 3.58 x 10⁴ x 4.184) / 1
x = 149787.2 joules
x = 1.5 x 10⁵ J
Answer:
Ba(ClO₃)₂ → BaCl₂ + 3 O₂ <em>Descomposition. </em>
CaO + CO₂ → CaCO₃ <em>Combination. </em>
NaNO₂ + HCl → NaCl + HNO₂ <em>Double replacement. </em>
Mg + ZnSO₄ → MgSO₄ + Zn <em>Single replacement. </em>
Explanation:
A combination reaction is defined as a reaction in which two or more substances combine to form a single new substance.
A + B → AB
A descomposition reaction is defined as a reaction in which a compound breaks down into two or more simpler substances.
AB → A + B
A double replacement is a chemical reaction were the positive and negative ions of two ionic compounds exchange places to form two new compounds.
AB + CD → AD + CB
A single replacement is another type of reaction were one element replaces a similar element in a compound.
A + BC → AC + B
Thus, with this information it is possible to classify these reactions as:
Ba(ClO₃)₂ → BaCl₂ + 3 O₂ <em>Descomposition. </em>One single molecule breaks down into two or more molecules
CaO + CO₂ → CaCO₃ <em>Combination. </em>Two substances are combined to form one single molecule
NaNO₂ + HCl → NaCl + HNO₂ <em>Double replacement. </em>Na and H are exchange places to form two new compounds
Mg + ZnSO₄ → MgSO₄ + Zn <em>Single replacement. </em>Mg is replacing Zn.
I hope it helps!
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38
Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62
Thus 62/100 are V+2
62/100 * 100 = 62%
</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
I need more information... what was the experiment??