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Amiraneli [1.4K]
3 years ago
12

Describe one way you could protect a bicycle from corrosion

Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
6 0
Don't use the brakes too hard when u stop. That can cause the biker to fall forward.

Give me a thanks if it helps!!
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Barium carbonate  barium oxide and carbon dioxide balanced formula HELP ME PLEASE
Sauron [17]

Answer:

4Ba(CO3) -> 4BaO2 + 2CO2

Explanation:

I looked at the oxygens to balance this. Ba(CO3) normally has 3 oxygens. BaO2 and CO2 have 4 oxygens total. The common multiple of 3 & 4 is 12. So there should be 12 oxygens on both sides. Then I just found the coefficients that would give 12 oxygens on both sides and can balance the rest of the atoms.

4 0
3 years ago
Jim can jog 800 m in 16 minutes. How many meters can he jog<br> in a minute?
RSB [31]

Answer: 50 meters

Explanation:

5 0
4 years ago
What is the pH of a 6.7 x 10^−5 M H+ solution?
lyudmila [28]

Answer:

pH = 4.17

Explanation:

According to the molar concentration you stated, pH of the solution is: 4.17

Remember that pH = - log [H⁺]

and [H⁺] = 10^-pH

When:

pH > 7 →  Basic solution

pH = 7 → Neutral solution

pH < 7 → Acid solution

3 0
4 years ago
Read 2 more answers
1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g
Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0
4 years ago
Help will give brainliest!!
GarryVolchara [31]

Half reaction :

Oxidation

Cu ⇒ Cu²⁺ + 2e⁻

Reduction

2Ag⁺+ 2e⁻ ⇒2Ag  

<h3>Further explanation</h3>

Given

Reaction

Cu+2AgNO₃⇒Cu(NO₃)₂+2Ag

Required

Oxidation and reduction half-reactions

Solution

Oxidation is an increase in oxidation number, while reduction is a decrease in oxidation number.

  • Oxidation

Cu ⇒ Cu²⁺ + 2e⁻

0 to +2

  • Reduction

2Ag⁺+ 2e⁻ ⇒2Ag  

+1 to 0

6 0
3 years ago
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