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(1) using newton's second law of motion force vecf=mveca=m (dvecv)/ (dt) where veca represents the acceleration of the particle and vecv its velocity Clearly, because of equal and opposite forces (newton's third law) they both feel the same force. In terms of acceleration equation (1) becomes (dvec v)/dt = q/m (vece+vecv×vecb).

I think we have to model these as charged conducting spheres so the charges are at the surface as shown Once you know how many moles you get in 0.742 g of gas, you can find its molar mass Gauss' law tells us that

Int int_s mathbf e cdot d mathbf s = (sum q_(enc))/epsilon_o and by using a concentric gaussian sphere, we can say of a general sphere of radius r that

Mathbf e = (sum q_(enc))/( 4 pi epsilon_o r^2) for these conducting spheres, we have 2 different situations. For 2 charges q_1 and q_2 separated by a distance r the force of attraction is given by coulomb's law (q_1q_2)/ (r^2) the constant (1)/ (4piepsilon_0) can be written as k and has the value 9xx10^ (9)m/f the work done in separating the 2. Lechatelier's principle => if a stress is applied to a reaction, the reaction will shift away from the applied stress and establish a new equilibrium

Three 'stress' factors affect stability of an equilibrium Changes in concentration of reactants or products, 2 Once the stopcock is opened, each gas will occupy the total volume of the two containers and will exert its own partial pressure independently of the other gas. The mass of water originally in the calorimeter was 76.58 g

We must identify the heat transfers that are occurring here

One is the heat transferred from the hot water as it cools (q_1) Another is the heat transferred to the water in the calorimeter as it warms (q_2) The third is the heat transferred to the calorimeter as it warms (q_3) Per the law of conservation of energy, the sum of the.

Here the mass of he gas w=11.28g the molar mass of he m_ (he)=4g/mol so number of moles of he in the sample n_ (he)=w/m_ (he)=11.28/4=2.82mol now initial volume of the gas v_1=63.2l initial pressure of the gas p_1=101.325kpa initial temperature of the gas t_1=273k final pressure of the gas p_2=98.1kpa final temperature of the gas t_2= (32.2+273)k=305.2k final volume of the gas v_2= → f = q→ v × → b newton's second law of motion states that the force on the particle is equal to the rate of change of its momentum ∴ → f = → P = q→ v × → b as momentum → p = m→ v, if we take the dot product of both sides with → p.

The idea here is that you need to use the ideal gas law to determine how many moles of gas you have in that sample

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