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svetoff [14.1K]
3 years ago
6

At 25 °C and 775 Torr, carbon dioxide has a solubility of 0.0347 M in water. What is its solubility at 25 °C and 1190 Torr? g

Chemistry
1 answer:
aniked [119]3 years ago
6 0

Answer:

g the day off and I will be there at work and I will be there at work and I will be there you go again with the person

Explanation:

no one is here yet but I can tomorrow if you have time can you come to my house and I will be there at work and I will be there at work and I will be there at work and I will be there

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I WILL GIVE BRAINLIEST ANSWER IN 5 MINS
snow_lady [41]

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.


i hope this part helps i will continue to research on Toothpaste's properties

6 0
3 years ago
Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta
AysviL [449]

<u>Answer:</u> The \Delta H_f 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:

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)

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)})]

We are given:

\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ

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

Hence, the \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

8 0
3 years ago
Is mass conserved when 50 g of sugar undergoes a physical change?
Lena [83]
Yes, mass never changes. No exceptions.
6 0
3 years ago
Read 2 more answers
100 POINTS PLEASE HELP (PHOTOS INCLUDED)
pshichka [43]

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

6 0
3 years ago
Read 2 more answers
Somebody answer please a picture is attached!??
DanielleElmas [232]
9.2 that is the answer
7 0
3 years ago
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