How many moles are in a 1.00-kg bottle of water? How many molecules? The molar mass of water is 18.0 g/mol.
时间: 2024-05-27 15:11:19 浏览: 7
To find the number of moles in a 1.00-kg bottle of water, we need to divide the mass by the molar mass:
1.00 kg / (18.0 g/mol) = 55.6 mol
Therefore, there are 55.6 moles of water in a 1.00-kg bottle.
To find the number of molecules, we need to use Avogadro's number, which is 6.022 x 10^23 molecules/mol:
55.6 mol x (6.022 x 10^23 molecules/mol) = 3.34 x 10^25 molecules
Therefore, there are 3.34 x 10^25 molecules of water in a 1.00-kg bottle.
相关问题
Calculate the needed amount of powdered reagents in order to prepare two liters of 1.50 M KOH Please show whole processes
Sure, I can help you with that! Here's the solution:
To calculate the needed amount of powdered reagents in order to prepare two liters of 1.50 M KOH, we need to use the following formula:
Number of moles of solute (n) = Molarity (M) x Volume of solution (V)
First, let's convert the given molarity into moles per liter:
1.50 M KOH = 1.50 moles of KOH per liter of solution
Next, we need to calculate the number of moles of KOH required to prepare two liters of the solution:
n = M x V
n = 1.50 moles/L x 2 L
n = 3.00 moles of KOH
Now we can calculate the mass of KOH we need using its molar mass and the number of moles:
Mass (in grams) = Number of moles (n) x Molar mass (Mw)
The molar mass of KOH is:
Atomic mass of K + Atomic mass of O + Atomic mass of H
39.10 g/mol + 15.99 g/mol + 1.01 g/mol
= 56.10 g/mol
Substituting the values:
Mass = 3.00 moles x 56.10 g/mol
Mass = 168.30 g
Therefore, we need 168.30 g of powdered KOH to prepare two liters of 1.50 M KOH solution.
Temp(C) Press(psi) Rate(C/min) DelP(N2) DelP(corr) #Moles Gas STP step vol chng (ml) Exo. Temp(C) Rate(C/min) 18650 Full Cell 24.834 11.0838 0.11 Sys Vol(non-vented) 11.08 0.00 0.00000 0.00 105.842 0.024 18.36 Cell ID 24.991 11.0544 0.32 11.93 11.09 -0.04 0.00000 -0.03 106.842 0.021 1827.9009 Cell ID-??g Elec. 25.291 11.4072 0.599 11.10 0.31 0.00001 0.23 107.843 0.02 18650 LCO 2.5Ah LGABHE21865 (P191GO96A2) (Lot #2) 0% SOC 25.497 11.1867 0.413 Hot (non-vented) 11.11 0.08 0.00000 0.06 108.849 0.016 25.667 11.1573 0.339 5 11.11 0.04 0.00000 0.03 109.853 0.015 25.801 10.9515 0.269 Hot (vented) 11.12 -0.17 -0.00001 -0.12 110.856 0.014 25.939 11.6277 0.275 13.6 11.12 0.50 0.00002 0.37 111.397 0.013 26.093 10.9956 0.309 11.13 -0.14 0.00000 -0.10 125.923 0.018 26.215 11.2455 0.243 Cold 11.14 0.11 0.00000 0.08 126.924 0.022 含义
这是一份数据表格,包括温度、压力、速率、气体量等参数。还涉及到一些电池的信息,如18650 Full Cell和18650 LCO 2.5Ah LGABHE21865等型号。数据表格中还提到了一些状态,如“Hot”和“Cold”,以及一些单位,如STP step vol chng(标准状况下体积变化)。这份数据表格可能是用于记录电池实验的结果和数据。