How many controls or controlled variables can there be in an experiment

What effect does an increase in carbon dioxide (CO 2 ) have on the concentration of hydrogen ions (H^ + ) in the blood?

Setler [38]

Answer:

As the level of carbon dioxide in the blood increases, more H+ is produced and the pH decreases. This increase in carbon dioxide and subsequent decrease in pH reduce the affinity of hemoglobin for oxygen. The oxygen dissociates from the Hb molecule, shifting the oxygen dissociation curve to the right.

4 0

3 years ago

Lysozyme extracted from chicken egg white has a molar mass of 13,930 g mo1-1. Exactly 0.1 g of this protein is dissolved in 50 g

Natalija [7]

Answer:

Relative lowering in vapor pressure of solution is 23.7599 mmHg.

Depression in freezing point of the solution is 0.0002667 K.

Elevation in boiling point of the solution is 0.0000745 K.

Osmotic pressure of the solution is 0.0035 atm.

Explanation:

1) Relative lowering in vapor pressure of solution containing non volatile solute is equal to the mole fraction of the solute.

$\frac{p^o-p_s}{p^o}=\frac{n_2}{n_1+n_2}$

$p^o$ = Vapor Pressure of the pure solvent

$p_s$ = Vapor Pressure of the solution

$n_2$ moles of solute

$n_1$ = moles of solvent

Given; $p^o=23.76 mmHg ,p_s=?$

Moles of Lysozyme = $n_2=\frac{0.1 g}{13,930 g/mol}=7.17\times 10^{-6} mol$

Moles of water = $n_1=\frac{50 g}{18 g/mol}=3.333 mol$

$\frac{23.76 mmHg-p_s}{23.76 mmHg}=\frac{7.17\times 10^{-6} mol}{3.333 mol+7.17\times 10^{-6} mol}$

$p_s=23.7599 mmHg$

2) Depression in freezing point $\Delta T_f$ is given by:

$\Delta T_f=K_f\times m$

$K_f$ = molal depression constant of solvent

m = molality of the solution = $\frac{moles}{\text{mass of solvent in kg}}$

$\Delta T_f=1.86 K kg/mol\times \frac{7.17\times 10^{-6} mol}{0.050 kg}$

$\Delta T_f=0.0002667 K$

3) Elevation in boiling point $\Delta T_b$ is given by:

$\Delta T_b=K_b\times m$

$K_b$ = molal elevation constant of solvent

m = molality of the solution = $\frac{moles}{\text{mass of solvent in kg}}$

$\Delta T_b=0.52 K kg/mol\times \frac{7.17\times 10^{-6} mol}{0.050 kg}$

$\Delta T_b=0.0000745 K$

4) Osmotic pressure of the solution $\pi$ is given as:

$\pi =cRT$

c = concentration of solution = $\frac{Moles}{Volume(L)}$

T = temperature of the solution

R = Universal gas constant = 0.0821 atm L/mol K

given , T = 298 K,

Mass of water = 50 g

Density of water = 1 g/ml

Volume of the water = $\frac{50 g}{1 g/mL}=50 mL=0.050 L$

$c=\frac{7.17\times 10^{-6} mol}{0.050 L}$

Since, there is less amount of solute in solvent volume of solution can be taken equal to the volume to the solution.

$\pi =\frac{7.17\times 10^{-6} mol}{0.050 L}\times 0.0821 atm L/mol K\times 298 K$

$\pi=0.0035 atm$

4 0

4 years ago

What type of ion does the substance give off if it turns the pH paper red?

ZanzabumX [31]

Answer:

Alkaline

Explanation:

3 0

4 years ago

Calculate the molarity of a solution of acetic acid made by dissolving 16.00 mL of glacial acetic acid at 25 ∘C in enough water

Soloha48 [4]

Answer : The molarity of solution is, 1.216 g/mole

Explanation : Given,

Density of acetic acid = 1.049 g/ml (standard value)

Volume of acetic acid = 16.00 ml

Volume of solution = 230.0 ml

Molar mass of acetic acid $CH_3COOH$ = $2(12)+4(1)+2(16)=60g/mole$

First we have to calculate the mass of acetic acid.

$Mass=Density\times Volume$

$Mass=1.049g/ml\times 16ml=16.784g$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of acetic acid}\times 1000}{\text{Molar mass of acetic acid}\times \text{volume of solution in ml}}$

$Molarity=\frac{16.784g\times 1000}{60g/mole\times 230ml}$

$Molarity=1.216g/mole$

Therefore, the molarity of solution is, 1.216 g/mole

7 0

3 years ago

I need help with this guys

Flura [38]

I have no idea I’m answering this because I need to ask more questions

5 0

4 years ago