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Tatiana [17]
3 years ago
11

Please help me fast ​

Chemistry
1 answer:
Mkey [24]3 years ago
7 0

Answer:

1..) The property by the virtue of which the metals can be beaten into sheets.

<u>Malleability</u><u>.</u>

<u>2</u><u>.</u>

<u>I</u><u>.</u><u>Reaction</u><u> </u><u>of</u><u> </u><u>iron</u><u> </u><u>with</u><u> </u><u>copper</u><u> </u><u>sulphate</u><u> </u><u>solution</u>

<u>ii</u><u>.</u>

<em>Iron metal displaces copper from its compound copper sulphate and forms iron sulphate along with copper metal. This reaction is a displacement reaction.</em>

<em>Reaction</em><em>:</em>

<u>Fe(Iron)</u>+CuSO4<u>(Copper Sulphate)</u><u>-</u><u>-</u><u>-</u><u>-</u><u>-</u><u>-</u><u>-</u><u>FeSO4(Ferrous </u><u>Sulphate</u><u>)</u><u>+</u><u>Cu(Copper)</u>

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Lithium, sodium and potassium are all in group 1 of the periodic table. Which of these has the lowest boiling point?
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Explanation:

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Thus, among the 3, potassium is furthermost down the group by virtue of the factor it has the highest atomic number.

Therefore, we can say that potassium has the lowest boiling point among the 3.

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Read 2 more answers
EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?
AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq).

The equilibrium constant for this reaction is called the self-ionization constant of water:

K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}].

Note that water isn't part of this constant.

The value of K_w at 25 °C is 10^{-14}. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}
  • However, [\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3} for pure water.
  • As a result, K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14} at 25 °C.

Back to this question. [\text{H}^{+}] is given. 25 °C implies that K_w = 10^{-14}. As a result,

\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
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