The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are,
= 9.25g/mL,
= 8.634g/mL and
= 9.130g/mL respectively.
The equation of the standard deviation is √{∑(x -
)÷N}
Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.
The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.
The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.
The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.
The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be
. Thus the standard deviation of the density value is 0.265g/mL.
The balanced reaction is:<span>
Mg (s) + 2HCl (aq)
--> MgCl 2 (aq) + H 2 (g)
We
are given the amount hydrochloric acid to be used for the reaction. This will
be the starting point of the calculation.
40.0 g
HCl ( 1 mol HCl / 36.46 g HCl)
(1 mol H2 / 2 mol HCl) (2.02 g H2 / 1 mol H2) = 1.11 g H2</span>
Answer:
1.605cm = 4 significant figures
16.050cm = 4 significant figures
16.050cm = 4 significant figures
12 + 12.5 + 125 = 149.5 = 4 significant figures
1.62 × 10^3/2.8 × 10^-5 = 1620/0.000028 = 11 significant figures
70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.
MM = 100.09 56.08
CaCO3 → CaO + CO2
Mass 125 g
a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)
= 1.249 mol CaCO3
b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)
= 1.249 mol CaO
c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g
The formula for average atomic mass is :
mass of isotope A * % of isotope A + mass of isotope B * % of isotope B + ....
Now,
Here,
Average atomic mass of nitrogen = (14.003 * 99.63%) + (15.000 * 0.37%)
= (14.003 * 0.9963) + ( 15.000 * 0.0037)
= 13.951 + 0.056
= 14.007 a.m.u.