Which of these changes would likely occur if the Earth's rotation on its axis

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 16.8g of carbon were burned in the presence of

Mrrafil [7]

61.6 grams of carbon dioxide was produced

4 0

2 years ago

When HClO2 is dissolved in water, it partially dissociates according to the equation

Natasha_Volkova [10]

Answer:

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Explanation:

Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

0.3473 = m * 1.86

Solving, m = 0.187 m

Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol

Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m

Initial molality

Assuming that a % x of the solute dissociates, we have the ICE table:

HClO2 H+ + ClO2-

initial concentration: 0.0854 0 0

final concentration: 0.0854(1-x/100) 0.0854x/100 0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

3 0

3 years ago

In which of the following sets is the symbol of the element, the number of protons, and the number of electrons given correctly?

sveta [45]

Answer:

A.

Explanation:

The number of the them will result in A.

7 0

2 years ago

grapevine trimming and other aromatics can be wrapped in aluminum foil with a few holes poked in it to?

Juli2301 [7.4K]

I need a little more context but I believe you are correct

7 0

2 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

2 years ago