If a aluminum rod was made out of solid glass instead of solid aluminum what would happen.

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(

Romashka [77]

<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

6 0

3 years ago

A student sets up the following equation to convert a measurement.

dybincka [34]

Answer:

$\frac{1 m}{100 cm}$

Explanation:

The final answer has a different set of units. In particular, meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since m is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportion:</u>

1 m = 100 cm

The full expression:

As you can see, the old unit (m) cancels out and you are left with cm in the denominator.

7 0

2 years ago

What is the Net Force?<br> 2N ><br> 4N -<br> SN

maria [59]

The net force is SN haha

5 0

4 years ago

If any girl want to be my girlfriénd?

Dafna1 [17]

Answer:

I can be a friend if u need

7 0

3 years ago

Read 2 more answers

What is the final pressure of a system (atm) that has the volume increased from 0.75 l to 1.1 l with an initial pressure of 1.25

lana66690 [7]

Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature
PV = k
where P - pressure , V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
1.25 atm x 0.75 L = P x 1.1 L
P = 0.85 atm
final pressure is B) 0.85 atm

3 0

3 years ago