Answer is 273
explanation:
The rule of division for significant figures is to put the answer to the least amount of significant figures presented in the problem.
0.0131 only has 3 sig figs which is 131
3.582 has 4 sig figs
The least amount is 3 in the problem
When you originally divide 3.582 and 0.0131 you get 273.4531 in your calculator
Now put it to 3 sig figs
4 is not greater than 5 so you don't need to round
273 is 3 sig figs.
So answer is 273
Answer:
δ N2(g) = 1.1825 g/L
Explanation:
- δ ≡ m/v
- Mw N2(g) = 28.0134 g/mol
ideal gas:
∴ P = (837 torr)×( atm/760 torr) = 1.1013 atm
∴ T = 45.0 °C + 273.15 = 318.15 K
∴ R = 0.082 atm.L/K.mol
⇒ n/V = P/R.T
⇒ n/V = (1.1013 atm) / ((0.082 atm.L/K.mol)(318.15 k))
⇒ n/V = 0.0422 mol/L
⇒ δ N2(g) = (0.042 mol/L)×(28.0134 g/mol) = 1.1825 g/L
1AlBr3+ 3K ---> 3KBr + 1Al
this is the answers and working out to both questions
I think that the answer is B, but I may be wrong...
