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Vladimir [108]
3 years ago
9

Help me please??????

Chemistry
2 answers:
snow_tiger [21]3 years ago
8 0

Answer:

a

Explanation:

Snezhnost [94]3 years ago
7 0

Answer:

It would be the first one

Explanation:

According the solubility rules if SO4 is with Pb then it is insoluble.

You might be interested in
5. How can you tell the difference between CuS and Cu2S
puteri [66]

Answer:

Density, melting point. and magnetic properties  

Explanation:

I can think of three ways.

1. Density

The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.

It should be possible to distinguish these even with high school equipment.

2. Melting point

Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).

A Bunsen burner can easily reach these temperatures.

3. Magnetic properties

You can use a Gouy balance to measure the magnetic susceptibilities.

In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.

In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.

A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.

A sample of CuS will be attracted by the magnetic field and show an increase in weight.

In the picture below, you can see the sample partially suspended between the poles of an electromagnet.

 

6 0
3 years ago
A group of students created a model showing a type of matter which of the following discribes what is shows by the model?
Lady_Fox [76]

Answer:

liquid

Explanation:

6 0
3 years ago
According to the ph scale, what is the approximate ph of ammonia solution?
UNO [17]
I believe that the answer is 11.5
5 0
3 years ago
300×175<br> Express your answer using the correct number of significant figures.
icang [17]

Answer:

300×175=52500

Explanation:

caculator

3 0
3 years ago
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
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