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The atoms in ceramic materials are held together by a chemical bond. The two most common chemical bonds for ceramic materials are covalent and ionic. For metals, the chemical bond is called the metallic bond. The bonding of atoms together is much stronger in covalent and ionic bonding than in metallic.
Answer:
pH=10.97
Explanation:
the solution of methyl amine with methylammonium chloride will make a buffer solution.
The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:
![pOH=pKb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
pH = 14- pOH
Let us calculate pOH
[Salt] = [methylammonium chloride] = 0.10 M (initial)
After adding base
![[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M](https://tex.z-dn.net/?f=%5Bsalt%5D%20%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X20%7D%7B%2820%2B50%29%7D%3D%200.0286M)
[base] = [Methylamine]=0.10
After mixing with salt
![[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M](https://tex.z-dn.net/?f=%5Bbase%5D%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X50%7D%7B%2820%2B50%29%7D%3D%200.0714M)
pKb= -log[Kb]= 3.43
Putting values
pOH = ![3.43+log(\frac{[0.0286]}{0.0714}](https://tex.z-dn.net/?f=3.43%2Blog%28%5Cfrac%7B%5B0.0286%5D%7D%7B0.0714%7D)
Answer:
A) 3.59 cm
Explanation:
Given that :-
The density of the gold ingot = 
Given that:- Mass = 5.50 lbs
Also, considering the conversion of lbs to g as shown below:-
1 lb = 453.592 g
Thus,
Mass = 
= 2494.756 g
The volume = Length*Breadth*Height
Given that:- Length = 12.0 cm , Breadth = 3.00 cm
Considering the expression for density as:-
Solving for height, we get that:-
Height=3.59 cm
Density H2O = 1g/cm³
1,5 kg H2O = 1500g = 1500cm³ (1dm³ = 1000cm³)
3moles of NaCl-----in---------1500cm³ H2O
x moles of NaCl ----in--------1000cm³ H2O
x = 2moles of NaCl
answer: 2 mol/dm³
Answer:
The answer is: 51.8 g (86% of serving size)
Explanation:
In order to solve the problem, we have to first determine the number of moles there are in 11.0 g of sucrose. Sucrose has a molecular weight of 342 g (we calculate this from the molar mass of the elements : 12 x 12 g/mol C + 22 x 1 g/mol H + 11 x 16 g/mol O). So, we divide the mass (11.0 g) into the molecular weight of sucrose:
11.0 g sucrose x 1 mol/342 g sucrose= 0.032 mol
We have 0.032 mol of sucrose in a serving of 60 g. But we need less moles (0.0278 mol):
0.032 mol ------------ 60 g serving
0.0278 mol------------ x= 0.0278 mol x 60 g serving/0.032 mol
x= 51.8 g
So, lesser than 1 serving of 60 g must be eaten to consume 0.0278 mol os sucrose. Exactly, 51.8 g (which stands for a 86% of the serving size).
