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adelina 88 [10]
4 years ago
10

Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.

Chemistry
1 answer:
tamaranim1 [39]4 years ago
3 0

Answer:

1500kg/m^3

Explanation:

Formula:

d=m/v

Given:

m=1.5kg

v=1000cm^3

(The side length of a cube is always equal to the others)

Required:

d=?

Solution:

d=m/v

d=1.5kg/1000cm^3

d=1.5kg/0.001m^3

d=1500kg/m^3

Hope this helps ;) ❤❤❤

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Theoretically, what mass of [Co(NH3)4(H2O)2]Cl2 could be produced from 4.00 g of CoCl2•6H2O starting material. If 1.20 g of [Co(
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<u>Answer:</u> The theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

<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 CoCl_2.6H_2O = 4.00 g

Molar mass of CoCl_2.6H_2O = 238 g/mol

Putting values in equation 1, we get:

\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol

The chemical equation for the reaction of CoCl_2.6H_2O to form  [Co(NH_3)_4(H_2O)_2]Cl_2 follows:

CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O

By Stoichiometry of the reaction:

1 mole of CoCl_2.6H_2O produces 1 mole of [Co(NH_3)_4(H_2O)_2]Cl_2

So, 0.0168 moles of CoCl_2.6H_2O will produce = \frac{1}{1}\times 0.0168=0.0168mol of [Co(NH_3)_4(H_2O)_2]Cl_2

Now, calculating the mass of [Co(NH_3)_4(H_2O)_2]Cl_2 from equation 1, we get:

Molar mass of [Co(NH_3)_4(H_2O)_2]Cl_2 = 234 g/mol

Moles of [Co(NH_3)_4(H_2O)_2]Cl_2 = 0.0168 moles

Putting values in equation 1, we get:

0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g

To calculate the percentage yield of [Co(NH_3)_4(H_2O)_2]Cl_2, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 1.20 g

Theoretical yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 3.93 g

Putting values in above equation, we get:

\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%

Hence, the theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

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