What is the kinetic energy of the ball precisely 4 seconds after it starts falling?

The reduction of ketone groups, such as those present on coq, produces alcohols. in contrast, oxidation of aldehydes produces wh

nevsk [136]

On oxidation of aldehydes produces carboxylic acid functional group.

The product of oxidation of aldehydes depends upon whether the reaction occur in acidic medium or alkaline condition.

If oxidation of aldehydes occurs under acidic condition the product is carboxylic acid but if oxidation of aldehydes occurs under alkaline condition then reduce as well as oxidized product obtained which is known as disproportional product.

The oxidation of aldehydes occur through potassium dichromate, potassium permanganate or many more. The oxidation of aldehydes in the presence of base is known as cannizzaro's reaction.

learn about oxidation

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7 0

2 years ago

What will happen to the litmus strips?

Nitella [24]

If the solution is BASIC than it will turn purple but if ACIDIC it will turn pink.

8 0

3 years ago

What element is this?

LiRa [457]

i think helium hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

8 0

3 years ago

When blood is donated, sodium oxalate solution is used to precipitate Ca2+, which triggers clotting. A 122.0−mL sample of blood

icang [17]

Answer:

[Ca2+] = 3.36 * 10^-8 M

Explanation:

Step 1: Data given

A 122.0−mL sample of blood contains 9.70 * 10^−5 g Ca2+/mL.

A technologist treats the sample with 100.0 mL of 0.1550 M Na2C2O4.

Ksp = 2.30 * 10^−9

Step 2: Calculate mass of Ca2+

Mass Ca2+ = 9.70 * 10^−5 g Ca2+/mL * 122 mL

Mass Ca2+ = 0.011834 grams

Step 3: Calculate moles of Ca2+

Moles Ca2+ = mass Ca2+ / molar mass Ca2+

Moles Ca2+ = 0.011834 grams / 40.078 g/mol

Moles Ca2+ = 2.95 *10^-4 moles = 0.000295

Step 4: Calculate moles of C2O4^2-

Moles C2O4^2- = Molarity * volume

Moles C2O4^2- = 0.1550 M * 0.100 L

Moles C2O4^2- = 0.0155 moles

Step 5: Calculate limiting reactant

Ca2+ is the limiting reactant. It will completely be consumed ( 0.000295 moles).

C2O4^2- is in excess. There will 0.000295 moles be consumed. There will remain 0.0155 - 0.000295 = 0.015205 moles of C2O4^2-

Step 6: Calculate total volume

Total volume = 122.0 mL + 100.0 mL = 222.0 mL = 0.222 L

Step 7: Calculate concentration CaC2O4

[CaC2O4] = 0.000295 mol / 0.222 L

[CaC2O4] = 0.00133 M

Step 8: Calculate concentration of C2O4^2-

[C2O4^2-] = 0.015205 mol / 0.222L

[C2O4^2-] = 0.0685 M

Step 9: Calculate [Ca2+]

Ksp = 2.3 * 10^-9 = [Ca2+] [C2O42-]

2.3 * 10^-9 = X * (X+0.0685)

X = [Ca2+] = 3.36 * 10^-8 M

6 0

3 years ago

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:

Butoxors [25]

Answer: The equilibrium concentration of ICl is 0.27 M

Explanation:

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

Initial concentration of iodine gas = $\frac{0.45}{2}=0.225M$

Initial concentration of chlorine gas = $\frac{0.45}{2}=0.225M$

For the given chemical equation:

$2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11$

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091$

Initial: 0.225 0.225

At eqllm: 0.225-x 0.225-x 2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[ICl]^2}{[Cl_2][I_2]}$

Putting values in above equation, we get:

$9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668$

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

6 0

3 years ago