Answer:
Beryllium (Be) : 9.01 g/mol
Silicon (Si) : 28.09 g/mol
Calcium (Ca) : 40.08 g/mol
Rhodium (Rh) : 102.91 g/mol
Explanation:
<span>2.51 grams
You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication:
19.16g * 0.131 = 2.50996g
Rounding to 3 significant figures gives 2.51 g.</span>
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams
<u>Explanation:</u>
To calculate the mass for given number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Actual moles of lithium chlorate = 9.45 moles
Molar mass of lithium chlorate = 90.4 g/mol
Putting values in above equation, we get:
![9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g](https://tex.z-dn.net/?f=9.45mol%3D%5Cfrac%7B%5Ctext%7BActual%20yield%20of%20lithium%20chlorate%7D%7D%7B90.4g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BActual%20yield%20of%20lithium%20chlorate%7D%3D%289.45mol%5Ctimes%2090.4g%2Fmol%29%3D854.28g)
To calculate the theoretical yield of lithium chlorate, we use the equation:
![\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BActual%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Actual yield of lithium chlorate = 854.28 g
Percentage yield of lithium chlorate = 81.0 %
Putting values in above equation, we get:
![81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g](https://tex.z-dn.net/?f=81%3D%5Cfrac%7B854.28g%7D%7B%5Ctext%7BTheoretical%20yield%20of%20lithium%20chlorate%7D%7D%5Ctimes%20100%5C%5C%5C%5C%5Ctext%7BTheoretical%20yield%20of%20lithium%20chlorate%7D%3D%5Cfrac%7B854.28%5Ctimes%20100%7D%7B81%7D%3D1054.67g)
Hence, the theoretical yield of the lithium chlorate is 1054.67 grams