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3 years ago

Under what circumstances might be important to know the activity tendencies of elements?

Chemistry

1 answer:

ad-work [718]3 years ago

3 0

Answer: It is very important to know the activity tendencies of the elements. The activity tendencies tells us about whether the element is reactive or not.

In the redox-reaction where there is a need to know the oxidizing agent and reducing agent, we can know it easily from the activity tendencies. The elements lying above the reactivity series are better reducing agents.

In the substitution reactions, the activity tendencies helps us to know which element will replace the other. The element lying above in the series will replace the element lying below it.

$MX+NY \rightarrow MY+NX$

where, N is an element that lies above in the reactivity series

M is an element that lies below in the reactivity series

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During which stage of the water cycle could water enter the atmosphere as a gas? A. transpiration B. precipitation C. accumulati

Fantom [35]

Answer: Transpiration---A

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3 years ago

H2SO4 + 2NaOH → 2H2O + Na So.

gayaneshka [121]

Answer:

2 mol H

Explanation:

For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.

1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)

= 2 mol H

Thus, we get 2 mol of H are needed to completely react 1 mol of NaOH.

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4 years ago

Anyway, which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

marissa [1.9K]

Nitrate ions $\text{NO}_3^{-}$ and potassium ions $\text{K}^{+}$ .

<h3>Explanation</h3>

Potassium dichromate undergoes a double-decomposition reaction with barium nitrate to produce barium dichromate and potassium nitrate. The reaction is possible due to the low solubility of barium dichromate that precipitate out of the solution readily after its production.

Start with the balanced chemical equation for this process:

$\text{K}_2\text{CrO}_4 \; (aq) + \text{Ba}(\text{NO}_3)_2 \; (aq) \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{KNO}_3 \; (aq)$

Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

$2 \; \text{K}^{+}\; (aq)+ 2\; \text{CrO}_4^{-} \; (aq) + \text{Ba}^{2+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq) \\ \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{K}^{+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq)$

$\text{K}^{+}$ and $\text{CrO}_4^{-}$ are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.

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3 years ago

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mylen [45]

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