The tungsten metal used for filaments in light bulbs is made by reaction of tungsten trioxide with hydrogen: wo3(s)+3h2(g)→w(s)+3h2o(g)
part a how many grams of tungsten trioxide must you start with to prepare 1.80 g of tungsten? (for wo3, mw = 231.8 amu.)
part b how many grams of hydrogen must you start with to prepare 1.80 g of tungsten?

A) 2.27 grams of tungsten trioxide must be needed to prepare 1.80 g of tungsten.

B)To prepare 1.80 g of tungsten we will need 0.0587 grams of hydrogen gas.

Explanation:

A) Mass of tungsten prepared = 1.80 g

Moles of tungsten =

According to reaction, 1 mol of tungsten is obtained from 1 mole of tungsten trioxide.

Then 0.009783 moles of tungsten will be obtained from:

of tungsten trioxide

Mass of 0.009783 moles of tungsten trioxide :

0.009783 mol × 231.8 g/mol = 2.27 g

2.27 grams of tungsten trioxide must be needed to prepare 1.80 g of tungsten.

B) According to reaction,1 mol of tungsten is produced by 3 moles of hydrogen gas

Then 0.009783 moles of tungsten are produced by :

Mass of 0.02935 moles of hydrogen gas:

0.02935 mol × 2 g/mol =0.0587 g

To prepare 1.80 g of tungsten we will need 0.0587 grams of hydrogen gas.

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey

your answer would be “dry”