The answer is (3) HClO. In the Cl2, chlorine has an oxidation number of zero. In HCl, the oxidation number is -1. In HClO2, the oxidation number is +3. In HClO, it is +1. You can calculate this by using O with oxidation number of -2 and H with +1.

8 0

3 years ago

According to the oxygen-hemoglobin dissociation curve, PO2 in the lungs of 100 mm Hg results in Hb being 98% saturated. At high

Ad libitum [116K]

Answer:

Hb would be 78.4% saturated.

Explanation:

This problem can be solved by using simple unitary method.

At 100 mm Hg pressure of oxygen, Hb is saturated by 98%

So, at 1 mm Hg pressure of oxygen, Hb is saturated by $\frac{98}{100}$ %

Hence, at 80 mm Hg pressure of oxygen, Hb is saturated by $\frac{98\times 80}{100}$ % or 78.4%

Therefore, at 80 mm Hg pressure of oxygen in the lungs, Hb would be 78.4% saturated.

4 0

3 years ago

A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t

andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

$\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)$

Where:

μ (l): is the chemical potential of 2-propanol in solution

μ° (l): is the chemical potential of pure 2-propanol

R: is the gas constant = 8.314 J K⁻¹ mol⁻¹

T: is the temperature = 82.3 °C = 355.3 K

x: is the mole fraction of 2-propanol = 0.41

$\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)$

$\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}$

$\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}$

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

I hope it helps you!

8 0

3 years ago