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Marysya12 [62]
3 years ago
7

When a car engine burns gasoline, the results of the reaction are similar to when cells burn glucose. both reactions release car

bon dioxide and water. in cells, the chemical energy in food is converted to atp and heat. in a moving car, the chemical energy in gasoline is converted to __________?
Chemistry
1 answer:
-Dominant- [34]3 years ago
3 0

the answer is heat. while a car is in idol, the tailpipe gets very hot, (motorcycle, car, etc.) this also produces h20 which you can see dripping out of the tailpipe.

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Cu+k.HNO3 --> Cu(No3)2+..........+...........
pentagon [3]

Answer:

Cu + HNO3 → Cu(NO3)2 + <u>H2O</u> + <u>NO2</u>

Cu + 4HNO3 → Cu(NO3)2 + <u>2H2O</u> + <u>2NO2 </u>(balanced equation)

4 0
2 years ago
Review the families and classification of elements in the periodic table. Based on this information, you would expect elements t
inn [45]
It is b because I need to answer my first question
6 0
3 years ago
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A motorcyclist completes a journey at an average speed of 65 mph
Andrews [41]
227.5

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8 0
2 years ago
Select the correct answer. which substance in this redox reaction is the oxidizing agent? cu 2agno3 → 2ag cu(no3)2 a. n b. agno3
lana66690 [7]

AgNO₃ will act as the oxidising agent.

<h3><u>For the given chemical equation:</u></h3>

Cu + 2AgNO₃ →  2Ag + Cu(NO₃)₂

Half reactions for the given chemical reaction:

<u>Reducing agent:</u>

Cu → Cu²⁺ + 2e⁻

Copper is a reducing agent because it is losing 2 electrons, which causes an oxidation process.

<u>Oxidising Agent</u>:

Ag⁺ + e⁻ → Ag

The silver ion undergoes a reduction process and is regarded as an oxidizing agent since it is acquiring one electron per atom.

Hence, AgNO₃ is considered as an oxidizing agent and therefore the correct answer is Option B.

<h3><u>Oxidising and Reducing agents</u></h3>
  • An oxidizing agent is a substance that reduces itself after oxidizing another material. It passes through a reduction process in which it obtains electrons and the substance's oxidation state is decreased.
  • A reducing agent is a chemical that oxidizes after reducing another material. It passes through an oxidation process in which it loses electrons and the substance's oxidation state increases.

To know more about the process of Oxidation and Reduction, refer to:

brainly.com/question/4222605

#SPJ4

8 0
8 months ago
The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.
olchik [2.2K]

Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
  • V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L.

Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

5 0
3 years ago
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