Answer:hemoglobin that has a high affinity for oxygen
Explanation:
Haemoglobin is the oxygen carrying pigment in blood. It performs this function because of the presence of iron at the center of the haemoglobin which coordinates reversibly with oxygen thereby aiding delivery of oxygen to cells. At high altitudes where air is thinner and the partial pressure of oxygen is lower than sea level, haemoglobin must develop a greater affinity for oxygen in order to carry the scarce oxygen to cells.
Answer:
[H2]eq = 0.0129 M
[F2]eq = 1.0129 M
[HF]eq = 0.9871 M
Explanation:
∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2
experiment:
∴ n H2 = 3.00 mol
∴ n F2 = 6.00 mol
∴ V sln = 3.00 L
⇒ [H2]i = 3.00 mol / 3.00 L = 1 M
⇒ [F2]i = 6.00 mol / 3.00 L = 2 M
[ ]i change [ ]eq
H2 1 1 - x 1 - x
F2 2 2 - x 2 - x
HF - x x
⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2
⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115
⇒ x² = (2 - 3x + x²)(115)
⇒ x² = 230 - 345x + 115x²
⇒ 0 = 230 - 345x + 114x²
⇒ x = 0.9871
equilibrium:
⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M
⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M
⇒ [HF] = x = 0.9871 M
Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Number of proton K=19
so, 42 - 19 =23
then the answer in 19 protons and 23 neutrons
