<span>1. MgBr2
Soluble.
Rule: all the binary compounds of the group 17 (different to F) with metals are solubles, except those formed with Ag, Hg (I) and Pb.
2. PbI2
Insoluble.
Rule: it is one of the exceptions stated in the rule above.
3. (NH4)2CO3
Soluble.
Rule: salts containing NH4(+) are soluble.
4. ZnSO4
Soluble
Rule: </span><span>This salt is not an exception to the rule that most sulfate salts are soluble. Important exceptions to this rule include BaSO4,
PbSO4, Ag2SO4 and SrSO4
5. Sr(OH)2
Soluble (slightly soluble).
Rule: </span><span>Hydroxide salts of Group II elements (Ca,
Sr, and Ba) are slightly soluble</span>
Li2O + H2O → 2LiOH
This is the answer
3rd blank write 2
<span>Answer: 8.15s
</span><span />
<span>Explanation:
</span><span />
<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />
<span>r = k [N]
</span><span />
<span>r = - d[N]/dt =
</span><span />
<span>=> -d[N]/dt = k [N]
</span><span />
<span>2) When you integrate you get:
</span><span />
<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />
<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />
<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>
Answer:
The correct answer is 0.206 moles
Explanation:
According to the given scenario, the calculation of the number of moles of ammonium chloride is available in the resulting solution is given below:
Given that
Amount of 
is 11.0 grams
And, the volume is 235 mL
Now the molar mass of is 53.49g/mol
So, the number of moles presented is
= 11.0 ÷ 53.49
= 0.206 moles
hence, the number of moles of ammonium chloride are available in the resulting solution is 0.206 moles
