Answer:
the mirror is 12 cm away from the image
Explanation:
; the image will be virtual and it will form at 12 cm distance behind the mirror.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
![Molarity=\frac{moles}{\text{Volume of solution(L)}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7Bmoles%7D%7B%5Ctext%7BVolume%20of%20solution%28L%29%7D%7D)
Moles of glucose = ![\frac{18.5 g}{180 g/mol}=0.1028 mol](https://tex.z-dn.net/?f=%5Cfrac%7B18.5%20g%7D%7B180%20g%2Fmol%7D%3D0.1028%20mol)
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = ![\frac{0.1028 mol}{0.1 L}=1.028 mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B0.1028%20mol%7D%7B0.1%20L%7D%3D1.028%20mol%2FL)
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = ![M_1=1.208 mol](https://tex.z-dn.net/?f=M_1%3D1.208%20mol)
Volume of the solution taken = ![V_1=30.0 mL](https://tex.z-dn.net/?f=V_1%3D30.0%20mL)
Molarity of the solution after dilution = ![M_2](https://tex.z-dn.net/?f=M_2)
Volume of the solution after dilution= ![V_2=0.500L = 500 mL](https://tex.z-dn.net/?f=V_2%3D0.500L%20%3D%20500%20mL)
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
![M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}](https://tex.z-dn.net/?f=M_2%3D%5Cfrac%7BM_1V_1%7D%7BV_2%7D%3D%5Cfrac%7B1.208%20mol%2FL%5Ctimes%2030.0%20mL%7D%7B500%20mL%7D)
![M_2=0.07248 mol/L](https://tex.z-dn.net/?f=M_2%3D0.07248%20mol%2FL)
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
![0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}](https://tex.z-dn.net/?f=0.07248%20mol%2FL%3D%5Cfrac%7B%5Ctext%7Bmoles%20of%20glucose%7D%7D%7B0.1%20L%7D)
Moles of glucose = ![0.07248 mol/L\times 0.1 L=0.007248 mol](https://tex.z-dn.net/?f=0.07248%20mol%2FL%5Ctimes%200.1%20L%3D0.007248%20mol)
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer:
2,4-di Ethyl-2-methylpentane
C10H22
3,4 dimethyl heptane
C9H20
Explanation:
D. The empirical formula and the molar mass
Answer:
P4(s) + 5 O2 (g)→ P4O10
Explanation:
If we desire to write a balanced chemical reaction equation, the rule of thumb is simple; the number of atoms of each element on the right hand side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation. Once this condition is satisfied, the reaction equation is said to be balanced.
As we can see, we need one mole of P4 and five moles of O2 to produce one mole of P4O10.