Answer:
295 mL
Explanation:
Given data:
Volume needed = ?
Molarity of solution = 0.193 M
Mass of salt = 7.90 g
Solution:
Molarity = number of moles of solute / volume in L
Number of moles of salt = mass/molar mass
Number of moles of salt = 7.90 g/ 138.205 g/mol
Number of moles = 0.057 mol
Volume needed:
Molarity = number of moles / volume in L
0.193 M = 0.057 mol / volume in L
Volume in L = 0.057 mol / 0.193 M
Volume in L = 0.295 L
L into mL
0.295 L × 1000 mL/1L
295 mL
Answer:
3.2 g O₂
Explanation:
To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
2 H₂O -----> 2 H₂ + 1 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
3.6 g H₂O 1 mole 1 mole O₂ 31.996 g
---------------- x --------------- x --------------------- x --------------- = 3.2 g O₂
18.014 g 2 moles H₂O 1 mole
Yes, half life is always given in units of time, hope this helps
<span>Answer:
(16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m
a.)
(3.9704 m) x (1.86 °C/m) = 7.38 °C change
0.00°C - 7.38 °C = - 7.38 °C
b.)
(3.9704 m) x (0.512 °C/m) = 2.03 °C change
100.00°C + 2.03 °C = 102.03 °C</span>
