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3 years ago

Describe the solubility of al(oh3 with respect to ph.

1 answer:

6 0

The solubility of Al(OH)3 with respect to pH is soluble at low pH, insoluble in pH-neutral solution, and soluble at high pH. We can express it as this:
Solubility equilibrium expression for Al(OH)3 is
Al(OH)3 <==> Al3+(aq) + 3OH-(aq) 
Initial 0 1.0x10^-3 M 
change +s +3s
Remember that solubility is used in chemistry to describe the properties of a solid compound that is mixed with and fully dissolves in a fluid without leaving any undissolved particles.

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43 Which element reacts spontaneously with 1.0M HCl (aq) at room temperature?

KonstantinChe [14]

The correct answer is (4) zinc~

According to the reactivity series of metals, only zinc react with HCl (aq) at room temperature.

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3 years ago

Which will not appear in the equilibrium constant expression for the reaction below?

alexira [117]

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4 years ago

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Which process is similar to binary fission?

elixir [45]

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Answer: d. mitosis

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3 years ago

How many solutions does the system have? zero one two three four infinitely many.

Ber [7]

Answer:

the system has infinitely many solutions.

Explanation:

The system is 2x + y = 1 and 4x + 2y = 2. Solutions to a system are the intersection points. Since these two lines are the same line they intersect everywhere. There are infinitely many solutions.

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3 years ago

a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine

iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction $\frac{a}{b}$

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

$\frac{15.39 g of gold}{2.77 g of chlorine}$

The proportion is the equal relationship that exists between two reasons and is represented by: $\frac{a}{b}=\frac{c}{d}$

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

$\frac{15.39 g of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}$

Solving for the mass of gold gives:

$mass of gold=1 g of chlorine*\frac{15.39 g of gold}{2.77 g of chlorine}$

mass of gold= 5.56 grams

So, there is 5.56 g of gold for every 1 g of chlorine

5 0

3 years ago