Answer:
C6H6
Explanation:
We can obtain the molecular formula from the empirical formula.
What we need do here is:
(CH)n = 78
The n shows the multiples of both element present in the actual compound. It can be seen that carbon and hydrogen have the same element ratio here. We then use the atomic masses of both elements to get the value of n. The atomic mass of carbon is 12 a.m.u while the atomic mass of hydrogen is 1 a.m.u
(1 + 12)n = 78
13n = 78
n = 78/13 = 6
The molecular formula is
(CH)n = (CH)6 = C6H6
Answer:
all the elements in the same period have the same valence electrons.
<span>11.3 kPa
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = Absolute temperature
We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon.
Atomic weight argon = 39.948
Atomic weight neon = 20.1797
Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol
Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol
Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol
Now take the ideal gas equation and solve for P, then substitute known values and solve.
PV = nRT
P = nRT/V
P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L
P = 113.8892033 L*kPa / 5.00 L
P = 22.77784066 kPa
Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles.
0.024777375 mol / 0.049809918 mol = 0.497438592
Now multiply by the pressure
0.497438592 * 22.77784066 kPa = 11.33057699 kPa
Round the result to 3 significant figures, giving 11.3 kPa</span>
43.8 has 3 significant figures and 1 decimal.
<h3 /><h3>What are significant figures?</h3>
The term significant figures refer to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation.
All zeros that occur between any two non-zero digits are significant. For example, 108.0097 contains seven significant digits. All zeros that are on the right of a decimal point and also to the left of a non-zero digit are never significant. For example, 0.00798 contained three significant digits.
Hence, 43.8 has 3 significant figures and 1 decimal.
Learn more about significant figures here:
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