What will you suggest to a person who is suffering from the problem of acidity after overeating?

Chemistry

2 answers:

Bumek [7]3 years ago

3 0

Answer:

If someone is suffering from the problem of acidity after overeating Baking soda solution would be suggested as a remedy as it is basic in nature, it neutralises excess acid in the stomach.

uranmaximum [27]3 years ago

3 0

If someone is suffering from the problem of acidity after overeating Baking soda solution would be suggested as a remedy as it is basic in nature, it neutralises excess acid in the stomach. Lemon juice, Vinegar, Baking soda solution.

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Determine the number of bonding electrons and the number of nonbonding electrons in the structure of cs2.

trapecia [35]

Carbon disulfide ( S=C=S) is made up of one carbon atom and two sulfur atom.
Carbon has 4 valence electrons while sulfur has 6 electrons.
During bonding all the four carbon valence electrons are used in bonding while each sulfur atom contributes two electrons two the bond formation.
Therefore; 4 + 2(2) = 8 electrons are used for bonding, while 8 electrons ( four from each sulfur atom) are not used for bond formation.

3 0

3 years ago

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How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .