Set up your units to algebraically cancel out the original unit, and end up with the units of the answer in the final numerator. We'll first convert to moles by using Avogadro's number, then the atomic mass of tungsten (W) from the periodic table, since the mass, in grams, of one mole of any element is the atomic mass number:
2.1x10∧24 atoms W x (1 mole/6.022x10∧23 atoms) x (183.84g/1 mole) = 641.089 grams
Since your given number has only two significant figures (2.1), your answer should also only express in 2 sig figs: 6.4x10∧2 grams
So.. what’s the question?
Comment on this what it is so I can help you
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
1) 2 reactants are given, so we have to find which of them is an excess reactant and which of them will react completely.
Convert gram of Na and H2O into moles.
M(Na) =23.0 g/mol
10.0 g Na*1 mol Na/23.0 g Na≈0.4349 mol Na
M(H2O) = 2*1.0 +16.0 = 18.0 g/mol
50.0 g H2O*1 mol H2O/18.0 g H2O =2.778 mol H2O
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
from equation 2 mol 2 mol
given 0.4349 mol 2.778 mol
We can see that H2O is an excess reactant, so we are going to find amount of H2 using Na.
2)
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
from equation 2 mol 1 mol
given 0.4349 mol x mol
x=(0.4349*1)/2 =0.21745 mol H2
3) mass of H2
0.21745 mol H2 * 2.0 g H2/1 mol ≈ 0.435 g H2
Answer : 0.435 g H2.
Answer:
The solubility of a substance can be increased by increasing the amount of solvent in a solution.
Explanation:
Two ways by which the rate of dissolving for substance can be increased.
1 By increasing the temperature of the solution: Increase in temperature will in turn increase the entropy or disorder or randomness of the solute molecules in a solution thus increasing the rate of dissolving the substance in that solution.
2 By increasing the volume of the solution : increase in the volume of the solution by adding more solvent can also increase the dissolving rate of a specific substance in a solution.
Answer:
no
it doesnot beacuse it is independent of it's original length......
