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3 years ago

Propose the reaction between zinc and hydrochloric acid. Balance the equation and find out how

Chemistry

1 answer:

Pachacha [2.7K]3 years ago

8 0

Answer:

When Zinc reacts with hydrochloric acid the following reaction takes place:

Zn+2HCl→ZnCl

2(g)

↑

Here 1 molecules of hydrochloric acid taking part in the reaction,the number of molecules of hydrogen produced in the reaction.

4molesofHCl×

2molesofHCl

1moleofH

=2molesofH

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How do i design a controlled experiment to appropriately test a hypothesis?

Virty [35]

Answer:

you need only one independent variable because if not, you wont know what factors have changed your experiment.

Explanation:

3 0

3 years ago

Calculate the molar mass of the acid.

JulijaS [17]

Citric acid has the molecular formula C6H8O7 so you can add the molar masses of the elements from the periodic table. C has a molar mass of 12.01 g/mol, H has 1.01 g/mol and O has 15.999 g/mol. Now you calculate the total molar mass= (6*12.01 + 8*1.01 + 7*15.999). This yields a molar weight of 192.124 g/mol (anhydrous)

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3 years ago

A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud

Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

= 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence, Mass = gcm-3 * cm3

Mass = g

The mass of pentane the student should weigh out is 28.17g

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2 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

Select the keyword or phrase that will best complete each sentence.

pishuonlain [190]

Answer:

1) acetylide

2) enol

3) aldehydes

4) tautomers

5) alkynes

6) Hydroboration

7) Keto

8) methyl ketones

Explanation:

Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.

Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.

The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.

3 0

3 years ago