Fill in the blankword bank (can use more than once): less, increases, decreases, greaterno2cl(g)+no(g)⇌nocl(g)+no2(g )1.disturbing the equilibrium by adding no2cl qc to a value than kc.2.to reach a new state of equilibrium, qc therefore , which means that the denominator of the expression for qc .to accomplish this, the concentration of reagents , and the concentration of products

Explanation:

If we have the following reaction at equilibrium:

                                          aA + bB ⇄ cC + dD

where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B,  C, and D. For the reaction at a particular temperature:

                       Kc=([C]^c *[D]^d)/([A]^a *[B]^b)

where Kc is the equilibrium constant, which holds that for a reversible reaction at equilibrium  and a constant temperature, a certain ratio of reactant and product concentrations has  a constant value, Kc (the equilibrium constant). Note that although the concentrations  may vary, as long as the reaction in in equilibrium and temperatura don't change, the value of K remains constant.

For reactions that have not reached equilibrium, we obtain the reaction quotient (Qc), instead of the equilibrium  constant by substituting the initial concentrations into the equilibrium constant expression.

                        Qc=([Co]^c *[Do]^d)/([Ao]^a *[Bo]^b)

To determine the direction in wich the net reaction will proceed to reach equilibrium, que compare the values of Qc and Kc.

Solution:

We have the following reaction:

                             NO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)

So:

Kc=([NOCl]^1*[NO2]^1)/([NO2Cl]^1 *[NO]^1)

     =([NOCl][NO2])/([NO2Cl][NO])  

1. In the equation above, [NO2Cl] it's in the denominator, so if we increase it's numericall value by adding NO2Cl decreases Qc  to a value less than Kc.

(From the chemical point of view, if we disturb the equilibrium adding NO2Cl (a reactant), to reach equilibrium again the system proceeds from left to right (→) consuming this reactant.)

2. To reach a new state of equilibrium (where Qc = Kc), Qc therefore  increases wich means that the denominator of the expression for Qc  decreases (in order to increase the denominator as mention above).

3. To accomplish this, the concentration of reageants decreases (reagents are being consumed), and the concentration of prodcuts increases (products are being formed).

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied