How can you test the complete separation of camphor and sand​

Response: sodium + water → sodium hydroxide + hydrogen

Clarification:

1) The initial part of the statement "sodium combines with water" indicates that these pure substances serve as the reactants.

Sodium belongs to the alkali metals (group 1: Li, Na, K, Rb, Cs, Fr).

A common characteristic of alkali metals is their reactive behavior with water, resulting in the formation of the respective hydroxide and hydrogen gas.

2) The following words, "to produce" suggest that what comes next represents the products of the reaction. In chemical equations, this is indicated by the arrow →

3) Finally, it specifies the products of the reaction: sodium hydroxide and hydrogen gas.

4) The word equation observes the sequence reactants → products, translates here as sodium + water → sodium hydroxide + hydrogen

5) Using chemical symbols and formats, this can be represented as:

Na(s) + H₂O (l) → NaOH(aq) + H₂(g).

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

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