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:
Al(OH)₃(aq) + 3 HBr(aq) ⇒ AlBr₃(aq) + 3 H₂O(l)
Explanation:
Let's consider the unbalanced equation for the reaction between solid aluminum hydroxide and a solution of hydrobromic acid. This is a neutralization reaction so it forms a salt, aluminum bromide, and water.
Al(OH)₃(aq) + HBr(aq) ⇒ AlBr₃(aq) + H₂O(l)
We start balancing Br atoms by multiplying HBr by 3.
Al(OH)₃(aq) + 3 HBr(aq) ⇒ AlBr₃(aq) + H₂O(l)
To get the balanced equation, we must multiply H₂O by 3 as well.
Al(OH)₃(aq) + 3 HBr(aq) ⇒ AlBr₃(aq) + 3 H₂O(l)
Answer:
Explanation:
Heat of vaporization is the amount of energy (enthalpy) that must be added to a liquid substance, to transform a quantity of that substance into a gas. Heat of Vaporization is also known as heat of evaporation
Answer: ![2MgI_2+Mn(SO_3)_2\rightarrow 2MgSO_3+MnI_4](https://tex.z-dn.net/?f=2MgI_2%2BMn%28SO_3%29_2%5Crightarrow%202MgSO_3%2BMnI_4)
Explanation:
Double displacement reaction: it is a chemical reaction in which the reactants exchanges their ions to form new compounds as a products.
All the reaction are example of double displacement reaction beside reaction where magnesium iodide is reacting with manganese(II) sulfate to give magnesium sulfate and manganese(IV) iodide .
![2MgI_2+Mn(SO_3)_2\rightarrow 2MgSO_3+MnI_4](https://tex.z-dn.net/?f=2MgI_2%2BMn%28SO_3%29_2%5Crightarrow%202MgSO_3%2BMnI_4)
In this reaction , charge on manganese have changed from 2+ to 4+. Manganese in getting oxidized. Example of an oxidation reaction. Hence, this reaction is not an example of double displacement reaction.
It’s the second one, crest.
Crest is a section of wavelength.