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3 years ago

if a layer of sandstone is in contact with a mass of granite that contain small pieces of the Sandstone which rock is older

1 answer:

6 0

The sandstone is older.in order for the granite to have been formed with sandstone in its precnese then it had to have already exsisted in that area..making it older.

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Please help

aev [14]

Answer:

Copper(II) sulphate – sodium hydroxide reaction

The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.

5 0

3 years ago

HELP ME THE WHOLE PAGE !!!! Someone help please

Deffense [45]

Answer:

In the image below

Explanation:

Hope this helps you :)

7 0

3 years ago

A scientist finds an unusual object and decides to study it. Which observation indicates that the object is a living thing?

Ksivusya [100]

B) It is made
Of cells

4 0

3 years ago

Read 2 more answers

Silver nitrate solution reacts with calcium chloride solution according to the equation: 2 AgNO3 + CaCl2 → Ca(NO3)2 + 2 AgCl All

HACTEHA [7]

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.

2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl

mass, g 6.97 6.39 ?

MW ,g/mol 169.87 110.98 143.32

mol =m/MW 0.10 0.06 0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

5 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago