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frozen [14]
2 years ago
15

A 0.0035 M aqueous solution of a particular compound has pH = 2.46. The compound is (A) a weak base (B) a weak acid (C) a strong

acid (D) a strong base (E) a salt
Chemistry
1 answer:
slava [35]2 years ago
7 0

Answer:

(a) A strong acid

Explanation:

We have given the pH of the solution is 2.46

pH=2.46  

So the concentration of H^+=10^{-pH}=10^{-2.46}=0.00346

solution having H+ concentration more than H^+=10^{-7} is acidic

Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution

Note-The concentration of H^+ decide the behavior of the solution that is, it is acidic or basic

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Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as
nikitadnepr [17]

The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

Moles of copper(II) nitrate = \frac{469 g}{187.5 g/mol}=2.5013 mol

Moles of sodium sulfide = \frac{156 g}{78 g/mol}=2 mol

According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

Then 2 moles of sodium sulfide will react with:

\frac{1}{1}\times 2mol= 2 mol of copper (II) nitrate

As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.

According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

\frac{2}{1}\times 2mol=4 mol sodium nitrate

Mass of 4 moles of sodium nitrate :

85 g/mol × 4 mol = 340 g

Theoretical yield of sodium nitrate = 340 g

The theoretical yield of sodium nitrate is 340 grams.

7 0
2 years ago
Read 2 more answers
According to the VSEPR model, the arrangement of electron pairs around NH3 and CH4 is A. the same, because in each case there ar
aleksklad [387]

Answer:

Option "B" is correct.

Explanation:

According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In NH_{3} and  CH_{4}, the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.  

3 0
2 years ago
In each pair of statements, choose the example that is more reliable.
TEA [102]

Answer:

"statement 2" for the first pair and "statement 1" for the second pair

Explanation:

3 0
1 year ago
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In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
2 years ago
How many moles of tungsten (W,183.85 g/lol are in 415 grams of tungsten?
vladimir1956 [14]

Given mass of tungsten, W = 415 g

Molar mass of tungsten, W = 183.85 g/mol

Calculating moles of tungsten from mass and molar mass:

415 g * \frac{1 mol}{183.85 g} = 2.26 mol W

7 0
2 years ago
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