How many grams are there in 1.77 x 10^22 atoms of beryllium

0.265 grams Be (3 sig. figs. per coefficient of given 1.77 x 10²² atoms Be)

Explanation:

Calculation Sequence: #atoms => #moles by dividing by Avogadro's Number (6.02 x10²³) => #grams (multiply #moles by Be formula weight - 9.012g/mol

Given 1.77 x 10²² atoms Be = 1.77 x 10²² atoms Be x 1 mole Be/6.02 x 10²³ atoms Be/mole Be x 9.012 g Be/mole Be = 0.26497 (calculator answer)

= 0.265 grams Be (3 sig. figs. per coefficient of given 1.77 x 10²² atoms Be)

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.