In 2006, parts of Texas experienced brush fires. The fires affected 203 of the 254 counties in

Chemistry

1 answer:

USPshnik [31]3 years ago

8 0

Answer: G

Explanation:

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What is an formed when liquid reactants

Gnesinka [82]

im not understanding the question, but if its a solid then the reactants are precipitate

5 0

2 years ago

YOU DO:

mihalych1998 [28]

Answer:

67.1%

Explanation:

Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:

<em>Moles Na₂CO₃ - 105.99g/mol-:</em>

6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.

As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:

0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃

And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):

0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.

And percent of NaHCO₃ in the sample is:

10.06g NaHCO₃ / 15g Sample * 100 =

7 0

3 years ago

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 $\rm H_2SO_4$ .