What is the expected mass (in kg) of a 10 over 5 b isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg a.1.67 × 10-26 kg b.1.67 × 1026 kg c.2.51 × 10-26 kg d.2.51 × 1026 kg

Problem #3 (ch. 1, problem 15)the ideal gas law provides one way to estimate the pressure exerted by a gas on a container. the law isí‘ťí‘ť=푛푛푛푛푛푛푉푉more accurate estimates can be made with the van der waals equationí‘ťí‘ť=푛푛푛푛푛푛푉푉â’푛푛푟푟â’푞푞푛푛2푉푉2where the term nb is a correction for the volume of the molecules and the term an2/v2is a correction for molecular attractions. the values of a and b depend on the type of gas. the gas constant is r, the absolutetemperature is t, the gas volume is v, and the number of moles of gas molecules is indicated by n. if n = 1 mol of an ideal gas were confined to a volume of v = 22.41 l at a temperature of 0â°c (273.2k), it would exert a pressure of 1 atm. in these units, r = 0.0826.for chlorine gas (cl2), a = 6.49 and b = 0.0562. compare the pressure estimates given by the ideal gas law and the van der waals equation for 1 mol of cl2 in 22.41 l at 273.2 k. what is the main cause of the difference in the two pressure estimates, the molecular volume or the molecular attractions?

Achange in the number of neutrons in an atom will change an blank . when the number of protons changes in an atom, a new element will form.

The reactions shown here can be combined to make the overall reaction c(s) + h2o(g) ⇌ co(g) + h2(g) by reversing some and/or dividing all the coefficients by a number. a. c(s) + o2(g) → co2(g) k=1.363×10^69 b. 2 h2(g) + o2(g) → 2 h2o(g) k=1.389×10^80 c. 2co(g) + o2 (g) → 2 co2(g) k=1.477×10^90