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nignag [31]
2 years ago
11

What is the hydroxide ion concentration of the lake described in question 3?

Chemistry
1 answer:
Sholpan [36]2 years ago
3 0
The hydroxide concentration is written as [OH-]. A useful equation is that pH + pOH = 14. Since the pH is 4, the pOH must be 10. To get [OH-], again take the neg. anti-log of both sides: - anti log (pOH-) = - antilog(10) ---> [OH-]=1 x 10^-10 M or 1 x 10^-10 moles [OH-] per liter solution (in this case, lake).
The hydroxide ion concentration describe in question 3 is 10-10 M
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Which of the following accurately represents the relationship between ceramic and metal?
evablogger [386]
Search Results
Featured snippet from the web
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.
4 0
2 years ago
What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c
olya-2409 [2.1K]

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]}

pH = 14- pOH

Let us calculate pOHpOH=3.43+ (-0.397)=3.03

pH=14-pOH=14-3.03=10.97

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M

[base] = [Methylamine]=0.10

After mixing with salt

[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M

pKb= -log[Kb]= 3.43

Putting values

pOH = 3.43+log(\frac{[0.0286]}{0.0714}

4 0
2 years ago
A gold ingot weighs 5.50 ibs. If the density of gold is 19.31 g/cm^3, and the length and width of the ingot are 12.0 cm and 3.00
Aliun [14]

Answer:

A) 3.59 cm

Explanation:

Given that :-

The density of the gold ingot = 19.31\ g/cm^3

Given that:- Mass = 5.50 lbs

Also, considering the conversion of lbs to g as shown below:-

1 lb = 453.592 g

Thus,

Mass = 5.50\times 453.592\ g = 2494.756 g

The volume = Length*Breadth*Height

Given that:- Length = 12.0 cm , Breadth = 3.00 cm

Considering the expression for density as:-

Density=\frac{Mass}{Volume}

19.31=\frac{2494.756}{12.0\times 3.00\times Height}

Solving for height, we get that:-

Height=3.59 cm

3 0
2 years ago
What is the molality of a solution in which 3.0 moles of NaCl is dissolved in 1.5 Kg of water?
Nesterboy [21]
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
5 0
2 years ago
29. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servi
adoni [48]

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).

3 0
2 years ago
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