<span>pKb = - log 1.8 x 10^-5 = 4.7
moles NH3 = 0.0750 L x 0.200 M=0.0150
moles HNO3 = 0.0270 L x 0.500 M= 0.0135
the net reaction is
NH3 + H+ = NH4+
moles NH3 in excess = 0.0150 - 0.0135 =0.0015
moles NH4+ formed = 0.0135
total volume = 75.0 + 27.0 = 102.0 mL = 0.102 L
[NH3]= 0.0015/ 0.102 L=0.0147 M
[NH4+] = 0.0135/ 0.102 L = 0.132 M
pOH = pKb + log [NH4+]/ [NH3]= 4.7 + log 0.132/ 0.0147=5.65
pH = 14 - pOH = 14 - 5.65 =8.35
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Answer:
The percentage is k 
%
Explanation:
From the question we are told that
The mass of the stibnite is 
The volume of KBrO3(aq) is 
The concentration of KBrO3(aq) is 
Now the balanced ionic equation for this reaction is
The number of moles of 
is
substituting values
from the reaction we see that 1 mole of reacts with 3 moles of 
so 0.003325 moles will react with x moles of
Therefore
Now the molar mass of is a constant with a values of 
Generally the mass of is mathematically represented as
substituting values
The percentage of Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as
k 
k %
1. Cost (fairly high).
2. Weather Dependent.
3. Solar Energy Storage Is Expensive.
4. Uses a Lot of Space.
5. Associated with Pollution.
Answer:
68 g
Explanation:
Molar mass (C10H16) = 10*12.0 g/mol + 16*1.0 g/mol = (120+16)g/mol =
= 136 g/mol
m (C10H16) = n(C10H16)*M(C10H16) = 0.5 mol*136 g/mol = 68 g
n(C10H16) - number of moles of C10H16
M(C10H16) - molar mass of C10H16
