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Pachacha [2.7K]
2 years ago
5

explain what you would do expect caesium astatide to look like .will it be soluble in water ?explain your reasoning​

Chemistry
1 answer:
son4ous [18]2 years ago
3 0

Answer:

it will not be soluble in water Becoz it can only be

separated by passing it through silver nitrate solution

Explanation:

i hope you understand

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The density of liquid Z is 0.9237 g/mL. A student masses a cup
Ulleksa [173]

Answer:

37.65mL

Explanation:

Given parameters:

density of liquid  Z = 0.9237g/mL

Mass of liquidZ + mass of cup = 50.7g

Mass of cup= 15.92g

Volume of liquid Z in cup=?

Solution:

 Density is the mass per unit volume of a substance. It is mathematically expressed as shown below:

    Density = \frac{mass}{volume}

To find the volume of liquid Z, we know the density of the liquid but we dont know the mass yet.

Mass of liquidZ = 50.7g - mass of cup = 50.7g - 15.92g = 34.98g

Therefore:

 Volume of liquidZ = \frac{mass of liquidZ}{density of liquidZ}

                               =  \frac{34.98}{0.9237}

                               = 37.65mL

3 0
2 years ago
Draw the product(s) obtained when benzoquinone is treated with excess butadiene. using wedges and dashes, indicate the stereoche
Alex17521 [72]

Answer:

The structure can be found on the attached documents

Explanation:

6 0
2 years ago
Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A
lesantik [10]

<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.

<u>Explanation:</u>

The chemical formula of carbonic acid is H_2CO_3. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

               H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)

The expression of first equilibrium constant equation follows:

Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}

The chemical equation for the second dissociation of carbonic acid follows:

               HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

6 0
2 years ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
labwork [276]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

5 0
2 years ago
A 0.12m solution of an acid that ionizes only slightly in solution would be termed _____.
Fittoniya [83]
A 0.12 M solution of an acid that only slightly ionizes in solution would be termed a weak acid. Weak acids are acids which do not completely dissociate in water. Thus lowering the presence of hydronium ions which measures the pH of an acid.  <span />
5 0
2 years ago
Read 2 more answers
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