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Novosadov [1.4K]
3 years ago
5

Sometimes clothes removed from the dryer cling together. What kinds of charges are on the clothes ?

Chemistry
1 answer:
inn [45]3 years ago
4 0

The answer is "opposite charges."

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astraxan [27]

Answer:A

Explanation:I hope I’m right

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2 years ago
Which of the following are nonelectrolytes? Hc2h302 (acetic acid) CH30H (method alcohol) h2So4 and c12h22011 (sucrose or table s
hoa [83]

Answer:

C₁₂H₂₂O₁₁ and CH₃OH

Explanation:

Sucrose and methyl alcohol are nonelectrolytes. They do not ionize or conduct a current in aqueous solution.

HC₂H₃O₂ is a weak electrolyte. It produces only a few ions and is a poor conductor of electricity in aqueous solution.

HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂⁻

H₂SO₄ is a strong electrolyte. Its first ionization is complete, so it is a good conductor of electricity in aqueous solution.

H₂SO₄ + H₂O ⟶ H₃O⁺ + HSO₄⁻

8 0
3 years ago
What tranports oxygen into the cells?​
Ivenika [448]
Hemoglobin

The protein inside (a) red blood cells that carries oxygen to cells and carbon dioxide to the lungs is (b) hemoglobin
3 0
3 years ago
Read 2 more answers
So my science teacher says if you put water in a cup and then put oil in it the oil is "more dense than water" witch means the o
laila [671]
The oil would sink, due to the density of it
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4 years ago
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For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)
dangina [55]

Answer:

14.4 g of CO_{2} can be produced.

Explanation:

Balanced equation: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

                                        Molar mass (g/mol)

                  C_{4}H_{10}                    58.12

                    O_{2}                         32

                  CO_{2}                       44.01

So, 9.60 g of C_{4}H_{10} = \frac{9.60}{58.12}mol=0.165mol

      17.0 g of O_{2} = \frac{17.0}{32}mol=0.531mol

According to balanced equation-

2 moles of C_{4}H_{10} produce 8 moles of CO_{2}

So, 0.165 moles of C_{4}H_{10} produce (\frac{8}{2}\times 0.165)mol of CO_{2}  or 0.660 moles of CO_{2}

13 moles of O_{2} produce 8 moles of CO_{2}

So, 0.531 moles of O_{2} produce (\frac{8}{13}\times 0.531)moles of CO_{2} or 0.327 moles of CO_{2}

As least number of moles of CO_{2} are produced from O_{2} therefore O_{2} is the limiting reagent.

So, maximum amount of CO_{2} that can be formed = 0.327 moles

                                                                              = (0.327\times 44.01)g

                                                                              = 14.4 g

3 0
3 years ago
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