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1 year ago

What is the molarity of a solution in which 250 grams of NaCl are dissolved in 2.9 L of water?

Chemistry

1 answer:

Maksim231197 [3]1 year ago

3 0

Answer:

About 1.48 M.

Explanation:

The formula for molarity is mol/L.

So firstly, you must find the amount of moles in 250 grams of NaCl.

I do this by using stoichiometry. First, I find how nany grams are in a single mole of NaCl. This is around 58.44 grams/mole. Now that I know this, I can now use a stoich table. (250 g NaCl * 1 mol NaCl / 58.44 g NaCl). I plug this into my calculator.

I get that 250 grams of NaCl is equal to about 4.28 moles.

Now I just plug into the formula!

4.28 moles/2.9 L = about 1.48

I've attached a picture of my personal notes below which shows work I have done in similar equations.

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Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

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As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

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6.0 g of copper was heated from 20 degree c to 90 degree c . How much energy was used to heat cu?

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Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

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Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

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Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$

We are given:

$K_1=K_b$

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Putting values in above equation, we get:

$K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}$

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