Physical Properties: Sodium bicarbonate is an odorless, white crystalline solid or fine powder. It has a slightly alkaline taste. Its density is 2.20 g mL-1 and it decomposes in temperatures above 50 ºC. The decomposition yields to sodium carbonate. It is highly soluble in water and poorly soluble in acetone and methanol. It is insoluble in ethanol.
Chemical Properties: Sodium bicarbonate is an amphoteric compounds, it means the compound has a character acids an basic at the same time. It is highly soluble in water, resulting in a slighty alkaline solution.
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<u>Answer:</u> The
for HCN (g) in the reaction is 135.1 kJ/mol.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ](https://tex.z-dn.net/?f=-870.8%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28-80.3%29%29%2B%283%5Ctimes%20%280%29%29%2B%282%5Ctimes%20%28-74.6%29%29%5D%5C%5C%5C%5C%5CDelta%20H_f_%7B%28HCN%29%7D%3D135.1kJ)
Hence, the
for HCN (g) in the reaction is 135.1 kJ/mol.
Yes, mass never changes. No exceptions.
Answer:
Explanation:
3.
Knowns: 100mL of solution; concentration of 0.7M
Unknown: number of moles
Equation: number of moles = volume * concentration
Plug and Chug: number of moles = 100/1000 * 0.7 = 0.07 mole
Final Answer: 0.07mole
2.
Knowns: 5.50L of solution; concentration of 0.400M
Unknown: number of moles
Equation: number of moles = volume * concentration
Plug and Chug: number of moles = 5.5 * 0.4 = 2.20 mole
Final Answer: 2.20 mole