Answer:
The O atom will tend to attract the electrons.
Explanation:
The electronegativity of O (3.5) is much higher than H (2.1), which means it is more likely to attract electrons. The higher the electronegativity, the more attractive.
Answer:
please where are the questions and also make your questions clear.
C. 14 protons and 14 electrons
Answer:
The value of
for this reaction at 1200 K is 4.066.
Explanation:
Partial pressure of water vapor at equilibrium = 
Partial pressure of hydrogen gas at equilibrium = 
Total pressure of the system at equilibrium P = 36.3 Torr
Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:



The expression of
is given by:


The value of
for this reaction at 1200 K is 4.066.
Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %