In an experiment, a student wants to increase the rate of a reaction that involves gases. Which change to the reactants would accomplish this?
Increase the volume to decrease pressure and to increase concentration.
O Increase the volume to increase pressure and to decrease concentration.
O Decrease the volume to decrease pressure and to increase concentration.
O Decrease the volume to increase pressure and to increase concentration.

Decrease the volume to increase pressure and to increase concentration.

Explanation:

Increasing the concentration will increase the number of collisions per second and therefore the rate of reaction.

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The last option: Decrease the volume to increase pressure and to increase concentration.

Explanation:

You can support your choice on basis of the collision theory.  According to the collision theory, chemical reactions happen when the molecules collide with each other, in the correct orientation and with enough kinetic energy to overcome the activation energy.  As a consequence, the larger the number of collisions the larger the rate of a reaction.  In the case of a reaction that involves gases, decreasing the volume will increase the pressure and the concentration (volume is inversely related to both the pressure and the concentration) causing the molecules to be closer to each other and to collide with higher frequency, this is you will expect more collisions, and so an increase on the rate of the reaction. That is expressed by the last choice: decrease the volume to increase pressure and to increase concentration.

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃. 

35. a insulated cylinder of solid gold (k 310 kw/m-k) of radius v2 m and height 5 m is heated at one end until the temperature at each point in the cylinder is given by w(r, y, z) (30 z2)(2-(r2+ y2). determine the rate of heat flow across each horizontal disk given by z 1, z 2, and z 3, identifying which has the greatest rate of heat flow across it.

explanation: