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vladimir2022 [97]
3 years ago
15

What are the physical states in which water can exist in the atmosphere? Give an example of how we can sense each state.

Chemistry
1 answer:
Sliva [168]3 years ago
3 0
Liquid - rain
Solid - snow
Gas - water vapour (when it evapourates from the lithosphere)
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Carbon diselenide (CSe2) is a liquid at room temperature. The normal boiling point is 125°C, and the melting point is –45.5°C. C
Tema [17]

Answer:

Explanation:  The strengths of the inter molecular forces varies as follows -

        CO_{2}< CS_{2} < CSe_{2}

The normal boiling point of CSe2 is 125°C and that of CS2 is 116°C, which explains the trend that as we move down the group, the boiling point of e compound increases as the size increases.

This usually happens because larger and heavier atoms have a tendency to exhibit greater inter molecular strengths due to the increase in size . As the size increases, the valence shell electrons move far away from the nucleus, thus has a greater tendency to attract the temporary dipoles.

And larger the inter molecular forces, more tightly the electrons will be held to each other and thus more thermal energy would be required to break the bonds between them.

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3 years ago
1. Of 100 animals on a farm, 20% are horses, 15 % are pigs, 30% are cows, and 35% are
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B. a circle graph

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7 0
3 years ago
When a 3.00 g 3.00 g sample of KBr KBr is dissolved in water in a calorimeter that has a total heat capacity of 1.36 kJ ⋅ K − 1
cupoosta [38]

Answer:

Molar heat of solution of KBr is 20.0kJ/mol

Explanation:

Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.

The dissolution of KBr is:

KBr → K⁺ + Br⁻

In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:

q = 1.36kJK⁻¹ × 0.370K

q = 0.5032kJ

Moles of KBr in 3.00g are:

3.00g × (1mol / 119g) = 0.0252moles

Thus, molar heat of solution of KBr is:

0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>

3 0
3 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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3 years ago
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