Answer:
C. Y & Z
Explanation:
V, W are imaginary metals here because their valence electrons are typically less than 4. X, Y, Z are non-metals and have higher valence electrons. Here, if V or W bind with X, Y, or Z we make ionic bond (because metal + non metal = ionic). But, if X binds with Y or Z or any combinations of any two of the three non-metals results in covalent bond (non metal + non metal = covalent).
Thus, Y and Z make covalent.
Answer:
<h3>the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Explanation:
Amount of HBr dissociated
2HBr(g) ⇆ H2(g) + Br2(g)
Initial Changes 2.15 0 0 (mol)
- 0.789 + 0.395 + 0.395 (mol)
At equilibrium 1.361 0.395 0.395 (mole)
Concentration 1.361 / 1 0.395 / 1 0.395 / 1
at equilibrium (mole/L)
![K_c=\frac{[H_2][Br_2]}{[HBr]^2} \\\\=\frac{(0.395)(0.395)}{(1.361)^2} \\\\=\frac{0.156025}{1.852321} \\\\=0.084](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%280.395%29%280.395%29%7D%7B%281.361%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B0.156025%7D%7B1.852321%7D%20%5C%5C%5C%5C%3D0.084)
<h3>Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Answer:
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Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr = 
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
( ideal gs equation)
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
Answer:
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