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mojhsa [17]
3 years ago
10

Solve for V2 when P1 = 1.0 atm V1 = 22.4 L and P2 = 4.0 atm This is a Boyle’s law gas problem

Chemistry
1 answer:
morpeh [17]3 years ago
3 0
P1/V1=P2/V2
1/22.4=4/x
X=4 multiple by 22.4
V2=89.6L
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An ion-dipole force is a type of intermolecular force in which forces of attraction or repulsion occur between neighboring ions, molecules or atoms.
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1What is ""Fremy's salt""? Write the molecular formula of Fremy's salt. Write the structure of the product obtained from phenol.
EastWind [94]

Answer and Explanation:

The Fremy salt is a chemical compound its chemical name is disodium nitrosodisuphonate

The molecular formula of the Fremy salt is K_2NO(SO_3)_2

When phenol is treated with Fremy salt in presence of water then benzoquinone is formed

C_2H_5OH\rightarrow C_6H_4O_2 ( in presence of Fremy salt and water)

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3 years ago
What is highly reactive to water and forms an ionic bond with chlorine? It’s very shiny.
34kurt
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3 years ago
The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)
EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = \frac{20.0 g}{395 g/mol}=0.05063 mol

2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

3 0
3 years ago
What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.
AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

pH=pK_{a} + log(\frac{[Base]}{[Acid]})

According to the given conditions, the equation will become as follow

pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

pH=6.86 + log(\frac{0.058}{0.042})

pH=7.00

5 0
4 years ago
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