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3 years ago

Contrast oxidation and reduction.

Chemistry

2 answers:

solmaris [256]3 years ago

7 0

Answer:

Oxidation involves the <u>loss</u> of elections while reduction involves the gain of electrons.

is the answer it is not gain wrong

Explanation:

Vesna [10]3 years ago

4 0

Option D is the answer

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A piston starts out with a volume of 3.000 L at 250.0 K and 800.0 torr. It is heated to 500.0 K and has a new volume of 4.000 L.

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3 years ago

Which list shows the phases of matter in order from the greatest average kinetic energy to the least average kinetic energy per

avanturin [10]

Answer:

Explanation:

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3 years ago

Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.

Shtirlitz [24]

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

$HCOOH\leftrightarrow H^++HCOO^-$

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

$\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}$

$1.8*10^-^4=\frac{x^2}{c-x}$

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

$1.8*10^-^4=\frac{x^2}{0.12}$

Let's solve this for x. Multiply both sides by 0.12

$2.16*10^-^5=x^2$

taking square root to both sides:

$x=0.00465$

Now, we have got the concentration of $[H^+] .$

$[H^+] = 0.00465 M$

We know that, $pH=-log[H^+]$

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

7 0

3 years ago

Read 2 more answers

44.90 g of reactants are placed in a beaker

sergiy2304 [10]

Answer:

D. -1882J

Explanation:

We can solve the energy released in a chemical reaction in an aqueous medium using the equation:

Q = -m*C*ΔT

<em>Where Q is energy (In J),</em>

<em>m is mass of water (45.00g)</em>

<em>C is specific heat of water (4.184J/g°C)</em>

<em>And ΔT is change in temperature (25.00°C - 15.00°C = 10.00°C)</em>

<em />

Replacing:

Q = -45.00*4.184J/g°C*10.00°C

Q = -1882J

Right answer is:

<em />

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3 years ago

An engine cylinder contains 175 mL of gas at a pressure of 1.0 atm. As the engine runs, it compresses the cylinder, reducing the

sukhopar [10]

This problem is providing the initial volume and pressure of a gas in an engine cylinder and asks for the final pressure once the volume of the gas has decreased due to a compression. At the end, the result turns out to be 11.7 atm.

<h3>Boyle's law</h3>

In chemistry, gas laws allow us to calculate pressure, volume, temperature or moles depending on a specified change and based on the concept and equation of the ideal gas, which derives the well-known gas laws; Boyle's, Charles', Gay-Lussac's and Avogadro's.

Thus, since this problem provides initial and final volume and initial pressure for us to calculate the final pressure, we understand we need to apply the Boyle's law as a directly proportional relationship between these two:

$P_1V_1=P_2V_2$

Thus, we solve for the final pressure, P2, to get:

$P_2=\frac{P_1V_1}{V_2}=\frac{1.0atm*175mL}{15mL}\\ \\ P_2=11.7atm$

Learn more about ideal gases: brainly.com/question/8711877

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3 years ago