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3 years ago

For which pair of cations would the addition of dilute hydrobromic acid precipitate one but not the other?

Chemistry

1 answer:

iogann1982 [59]3 years ago

6 0

Answer:

a. Ag+ and Ca2+

The pair of cations related to the question are as below:

a. Ag+ and Ca2+

b. Hg2^2+ and Ag^+

c. Ba^2+ and Na^+

d. Ca^2+ and Ba^2+

e. Pb^2+ and Ag^+

Explanation:

a. Ag+ and Ca2+ correct

Ag+ will precipitate with dilute hydrobromic acid (HBr) but Ca+2 will not

Ag+ + Br- -----------------> AgBr (s)

Ca+2 + 2Br- ---------------> CaBr2 (aq)

B) Hg2^2+ and Ag^+ Incorrect

Both Hg2^2+ and Ag^+ will precipitate. One has to precipitate while the other does not, so this is incorrect.

C) Ba2 and Na+ both won't precipitate. Incorrect

One has to precipitate while the other does not, so this is incorrect.

D) Ca2+ and Ba 2+both won't precipitate. Incorrect

One has to precipitate while the other does not, so this is incorrect.

E ) Both Pb2 and Ag+ will precipitate. Incorrect

One has to precipitate while the other does not, so this is incorrect.

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a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter

ladessa [460]

Answer:

The concentration of chloride ion is $2.82\times10^{-3}\;mol/L$

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the $Cl^-$ content 100 ppm suggests the presence of 100 mg of $Cl^-$ in 1 L of solution.

The molar mass of $Cl^-$ is equal to the molar mass of Cl atom as the mass of the excess electron in $Cl^-$ is negligible as compared to the mass of Cl atom.

So, the molar mass of $Cl^-$ is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of $Cl^-$ present in 100 mg (0.100 g) of $Cl^-$ is calculated as shown below:

$N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol$

So, there is $2.82\times10^{-3}\;mol$ of $Cl^-$ present in 1 L of solution.

5 0

4 years ago

An unknown substance has the composition of 77.87% c, 11.76% h and 10.37% o. the compound has a molar mass of 154.25 g/mole. wha

frutty [35]

We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound
C H O
mass 77.87 g 11.76 g 10.37 g
number of moles 77.87/12 11.76/1 10.37/16
moles = 6.48 = 11.76 =0.648
divide by least number of moles
6.48/0.648 11.76/0.648 0.648/0.648
= 10 =18.1 = 1
rounded off
C - 10 , H - 18 and O - 1
empirical formula - C₁₀H₁₈O
mass of empirical unit = 12 x 10 + 1x 18 + 16 = 120 + 18 + 16 = 154
number of empirical units = molecular mass / mass of one empirical unit
= 154.25 / 154 = 1.00
Therefore molecular formula = C₁₀H₁₈O

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3 years ago

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Alex17521 [72]

Answer:

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Explanation:

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3 years ago

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poizon [28]

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3 years ago

Please help!!

andriy [413]

(I know this is late so hopefully other people find it helpful)

Answer: Solid Cu

Since this is a voltaic cell:

Copper is the cathode, therefore having a positive charge.

Zinc is the anode, therefore having a negative charge.

(Also, I took the exam and it's correct; good luck everyone!)

3 0

2 years ago