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storchak [24]
3 years ago
12

Which of the following best explains why the volume of an inflated balloon

Chemistry
1 answer:
mariarad [96]3 years ago
8 0
Yes it’s for the good of ways
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Why will paper stick to a ballon that you have rubbed on your hair?
Usimov [2.4K]
Depending on the balloon, the surface will cause friction between the two and oxygen molecules will fuse together to create a sticky sensation
7 0
4 years ago
Can any person help me for this problem? Anyone's welcome
noname [10]

Answer:

The anode made of the impure copper

The cathode made of pure copper

The electrolyte of copper (II) sulfate CuSO₄ solution

The silver impurities at the anode due to their high tendency of accepting electrons and being a stronger reducing agent than either copper or zinc will remain relatively in place and relatively stable and will not actively take part in the oxidation reaction taking place at the anode

The zinc impurities will be the first element of the three metals to give up electrons and go into the solution as zinc ions due to their high tendency to loan out two electrons and become oxidized into Zn²⁺ ions

The drawing of the electrolytic cell created with Microsoft Visio is attached

Explanation:

5 0
3 years ago
A certain element has a melting point over 700 ∘c and a density less than 2.00 g/cm3. give one possible identity for this elemen
Hunter-Best [27]

A melting point of over 700 C and a density of less than 2 g/cm3 can be observed for many group 2 elements. In this group, the density increases on moving down the group, whereas the melting point increases upto calcium and then starts decreasing.

Calcium, symbol Ca is the element with melting point around 840 C and density of 1.55 g/cm3 which is closest to the specified data range .

3 0
4 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
What does the speed of an object tell you about that object's motion?
Ugo [173]

Answer:

Your answer should be 3!

Explanation:

Hope this helps!!

6 0
3 years ago
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