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zalisa [80]
3 years ago
5

A certain material has a mass of 12.48 g while occupying 12.48 cm3 of space. What is this material?

Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
5 0
This material has a density of 1 g/cm3.
(since 12.48 g/ 12.48 cm3 = 1 g/cm3)

Therefore, this material is water.
Sliva [168]3 years ago
5 0

Answer: The material is water as it has density of 1g/cm^3  

Explanation:

Density is defined as the mass contained per unit volume.  It is characteristic of a material.

Density=\frac{mass}{Volume}

Given : Mass of object =12.48 grams

Volume of the object = 12.48cm^3

Putting in the values we get:

Density=\frac{12.48g}{12.48cm^3}

Density=1g/cm^3

Thus the density of the material is 1g/cm^3 and material is water which has density of  1g/cm^3.

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3 years ago
What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?
AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol

The given chemical equation follows:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = \frac{3}{2}\times 0.467=0.7005mol of copper

Now, calculating the mass of copper  from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g

Hence, the theoretical yield of copper is 44.48 grams

4 0
4 years ago
You are on an alien planet where the names for substances and the units of measures are very unfamiliar.
enyata [817]

Given that

1 skvarnick = 45 quibs

3 quibs = 7 sleps

Now if we have 45 quibs it means we have you have one skvarnick

three quibs = 7 sleps

so one quib = 7/3 sleps

so 45 quibs = 7 X 45 / 3 = 105 sleps

8 0
3 years ago
Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0
3 years ago
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