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3 years ago

A cough sytup contains 0.5M dextromethophan. How many moles of the cough supressant are in 21.3mL of the cough syrup?

Chemistry

1 answer:

Doss [256]3 years ago

7 0

Answer:

0.0107 mol

Explanation:

Multiply concentration by volume (in liters) to get moles.

0.5 M • 0.0213 L = 0.0107 mol

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Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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3 years ago

Two copper electrodes dipped in copper sulphate solution are connected to the end of the battery marked with a ''_'' is:

guajiro [1.7K]

Answer:

d) catión

Explanation:

8 0

3 years ago

Read 2 more answers

The atomic mass of an imaginary element X is 30.7377 amu. Element X has three isotopes. One isotope has an isotopic mass of 30.5

ryzh [129]

Answer:

wuhwwisbsjsbsisjsbsjsjiw

Explanation:

bhajwjbw wjsjw bwuebe e

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3 years ago

A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ions as

grigory [225]

Answer:

259.497 mg, 58.84%

Explanation:

BaSO₄ → Ba²⁺ + SO₄²⁻

to calculate the mole of BaSO₄

mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol

comparing the mole ratio

1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺

403 mg BaSO₄ yields ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺

441 mg BaSO₄ will yield ( 1.7268 × 137.327 × 441 mg ) / 403 mg = 259 .497 mg

mas percentage of the Barium compound = 259 .497 mg / 441 mg × 100 = 58.84%

8 0

3 years ago

What is 'Lo Salt' used for?

Igoryamba

It offers a route to reducing sodium intake.

Hope I helped.

8 0

3 years ago