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AleksAgata [21]
3 years ago
14

A student is determining the density of an unknown metal with a mass of 32.56 g. The student partially fills a graduated cylinde

r with water and measures the volume of the water by itself as 14.78 mL. The student then adds the metal to the water and measures the new volume as 20.44 mL. What is the identity of the metal
Chemistry
1 answer:
xenn [34]3 years ago
4 0

Answer:

The metal is Tin (Sn)

Explanation:

The following data were obtained from the question:

Mass of metal = 32.56 g

Volume of water = 14.78 mL

Volume of water + metal = 20.44 mL

Next, we shall determine the volume of the metal. This is illustrated below:

Volume of water = 14.78 mL

Volume of water + metal = 20.44 mL

Volume of metal =..?

Volume of metal = (Volume of water + metal) – (Volume of water )

Volume of metal = 20.44 – 14.78

Volume of metal = 5.66mL

Next, we shall determine the density of the metal.

Density of a substance is defined as the mass of the substance per unit volume of the substance. Mathematically, the density of a substance is expressed as:

Density = Mass / volume

With the above formula, we can obtain the density of the metal as follow:

Mass of metal = 32.56 g

Volume of metal = 5.66mL

Density =.?

Density = Mass /volume

Density = 32.56g/5.66mL

Density of the metal = 5.75g/mL

Comparing the density of the metal with standard density of elements, the metal is Tin (Sn)

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Assign an oxidation number to each atom in the products.
Rzqust [24]
<h3>Answer:</h3>

Oxidation number refers to a number given to an atom which shows the number of number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound.

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8 0
3 years ago
Read 2 more answers
At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.600 M . N 2 ( g )
alexandr1967 [171]

Answer:

0.78 M

Explanation:

First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:

N2(g) + O2(g) <------> 2NO(g)   Kc = ?

Writting Kc:

Kc = [NO]² / [N2] * [O2]

Replacing the given values we have then:

Kc = (0.6)² / (0.2)*(0.2)

Kc = 9

Now that we have the Kc, let's see what happens next.

We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:

      N2(g) + O2(g) <------> 2NO(g)

I:      0.2        0.2                 0.9

C:     +x         +x                   -2x

E:    0.2+x    0.2+x             0.9-2x

Replacing in the Kc expression we have:

Kc =  [NO]² / [N2] * [O2]

9 = (0.9-2x)² / (0.2+x)*(0.2+x)   ----> (this can be expressed as 0.2+x)²

Here, we solve for x:

9 = (0.9-2x)² / (0.2+x)²

√9 = (0.9-2x) / (0.2+x)

3(0.2+x) = 0.9-2x

0.6 + 3x = 0.9 - 2x

3x + 2x = 0.9 - 0.6

5x = 0.3

x = 0.06 M

This means that the final concentration of NO will be:

[NO] = 0.9 - (2*0.06)

[NO] = 0.78 M

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