Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Answer:
the number of protons and neutrons at both terminals are equal
Explanation:
When the number of positive charge and negative charge are both equally the terminal, it becomes neutral and out of charge, because first it undergo enough chemical reaction and there is no remaining tendency for positive and negative charges to get separated. When this tendency dies, the battery also will run out of charge.
Answer:
24.7 amu
Explanation:
An isotope is when an element can have different number of neutrons but they have same number of protons.
In order to calculate the average atomic mass with the given information do the following operations:
First change de percentages to fractional numbers, divide by 100.
I like to make a table, to organize all data and I believe is easier to understand.
65/100 = 0.65
35/100 = 0.35
% fraction
65.0 0.65
35.0 0.35
total100.0 1
Now multiply each mass with their corresponding fraction
24 (0.65) = 15.6
26 (0.35) = 9.1
% fraction uma uma
65.0 0.65 24 15.6
35.0 0.35 26 9.1
total100.0 1 24.7
Finally you add the resulting mass and the units will be in uma.
15.6+9.1 = 24.7
Therefore the average atomic mass of this element will be 24.7 uma.
Check the table in the document attached
Given:
P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?
Assume that the gas ideally, thus, we can use the ideal gas equation:
PV = nRT
where R = 0.0821 L atm/mol K
123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T
solve for T
T = 295.72 K<span />
