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patriot [66]
3 years ago
5

Which of these particles are lost in the oxidation process? A. protons B. neutrons C. electrons D. atoms

Chemistry
2 answers:
vladimir2022 [97]3 years ago
7 0

C. electrons

Hope this helps.

irina [24]3 years ago
3 0

C. electrons

i took at test that had this question and this was the answer

brainliest plz

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If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required ?
Oksi-84 [34.3K]

M_{A}V_{A}=M_{B}V_{B}\\(80.0)(4.00)=V_{B}(12.4)\\V_{B}=\frac{(80.0)(4.00)}{12.4} \approx \boxed{25.8 \text{ mL}}

5 0
2 years ago
Does the mass of water increase or decrease when it changes to ice
NikAS [45]

Answer:

The mass stays the same only volume changes, the volume decreases

Explanation:

The ice shrinks (decreases volume) and becomes more dense. The weight will not (and cannot) change.

4 0
3 years ago
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Select the situation below that would produce a total displacement of zero
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Answer:

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3 years ago
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If 4000 g of Fe2O3 reacts, how many moles of CO are needed?<br> (add work)
7nadin3 [17]

Answer:

75.15 mol.

Explanation:

  • Firstly, we need to write the balanced equation of the reaction:

<em>Fe₂O₃ + 3CO → 2Fe + 3CO₂.</em>

It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.

∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.

  • we need to calculate the no. of moles of (4000 g) of Fe₂O₃:

<em>no. of moles of Fe₂O₃ = mass/molar mass</em> = (4000 g)/(159.69 g/mol) = <em>25.05 mol.</em>

<u>Using cross multiplication:</u>

1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,

∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.

<em>∴ The no. of moles of CO needed</em> = (3.0 mol)(25.05 mol)/(1.0 mol) =<em> 75.15 mol.</em>

5 0
3 years ago
Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure
o-na [289]

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

6 0
3 years ago
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