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Tatiana [17]
3 years ago
9

If you are given the number of molecules in an unidentified chemical compound, can you calculate the number of moles in this sam

ple?
yes, because you know that each mole contains 6.02 x 1023 molecules

yes, as long as you can also measure the mass of the sample

no, because the number of moles depends on the volume of the sample

no, because the number of moles depends on the identity of the substance
Chemistry
2 answers:
natta225 [31]3 years ago
6 0
3rd one i hope that helps

sasho [114]3 years ago
6 0

Answer:

The correct option is A

Explanation:

Avogadro's number (6.02 × 10²³) is used to calculate the actual number of ions/atoms/moles present in the molecules of a chemical compound. To do that, avogadro's number is multiplied by the number of molecules of the compound - this is because the number of atoms/ions/moles present in one mole of any substance/compound is 6.02 × 10²³. Hence, one can calculate the the number of moles in the sample because the avogadro's number and the number of molecules of the unknown sample is known.

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A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
givi [52]

Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

V_{V^{+3}} = 59.0 \ mL

V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
The painkiller, Advil® contains the active ingredient ibuprofen (IB), which has a pKb of
denis23 [38]

This problem is providing the basic dissociation constant of ibuprofen (IB) as 5.20, its pH as 8.20 and is requiring the equilibrium concentration of the aforementioned drug by giving the chemical equation at equilibrium it takes place. The obtained result turned out to be D) 4.0 × 10−7 M, according to the following work:

First of all, we set up an equilibrium expression for the given chemical equation at equilibrium, in which water is omitted for it is liquid and just aqueous species are allowed to be included:

Kb=\frac{[IBH^+][OH^-]}{[IB]}

Next, we calculate the concentration of hydroxide ions and the Kb due to the fact that both the pH and pKb were given:

pOH=14-8.20=5.80

[OH^-]=10^{-5.8}=1.585x10^{-6}M

Kb=10^{-5.20}=6.31x10^{-6}

Then, since the concentration of these ions equal that of the conjugated acid of the ibuprofen (IBH⁺), we can plug in these and the Kb to obtain:

6.31x10^{-6}=\frac{(1.585x10^{-6})(1.585x10^{-6})}{[IB]}

Finally, we solve for the equilibrium concentration of ibuprofen:

[IB]=\frac{(1.585x10^{-6})(1.585x10^{-6})}{6.31x10^{-6}}=4.0x10^{-7}

Learn more:

(Weak base equilibrium calculation) brainly.com/question/9426156

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