The temperature of your water was 22.4 degrees Celsius. The volume of hydrogen collected was 37.1 mL. The atmospheric pressure in the lab room was 29.18 inches Hg. The difference in the water level between the beaker and the burette is 18.4 cm. What was the mass of the magnesium ribbon used?

The correct answer is 27.22 mg.

Explanation:

Based on the given information, the temperature of the water given is 22.4 degree C or 22.4 + 273 = 295.4 K

The volume of the gathered hydrogen is 37.1 ml or 0.0371 L

The pressure mentioned in the lab room is 29.18 in Hg or 29.18 × 2.54 = 74.1172 cm Hg.

18.4 cm is the difference given in the water level between the burette and the beaker, so the pressure is,

74.1172 cm - 18.4 cm = 55.7172 cm Hg or 557.172 mmHg

760 mmHg is equal to 1 atm, therefore, 557.172 mmHg will be equal to,

557.172 mm Hg / 760 mm Hg × 1 atm = 0.733 atm

The moles of hydrogen gas can be calculated by using the ideal gas equation, that is, PV = nRT

n = PV / RT

Putting the values we get,

n = 0.733 atm × 0.0371 L / 0.0821 atm L mol⁻¹ K⁻¹ × 295.4 K

n = 0.00112 mol

Let us consider the equation,

2 HCl + Mg ⇒ H₂ + MgCl₂

By seeing the reaction it is clear that for the generation of 1 mole of hydrogen gas, 1 mole of magnesium is needed.

The moles of magnesium is 0.00112 moles and the molecular mass of magnesium is 24.305 gram per mol

The mass of magnesium ribbon can be determined by using the formula, mass = mole × molecular mass

mass = 0.00112 mole × 24.305 g/mol

mass = 0.02722 grams or 27.22 mg.

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.