Anyone who has a general idea of how to do these

6. slopes of perpendicular lines are negative reciprocals of each other. Therefore if CD is , the slope of AB will be -

7. Note that the y intercept of the line is (0,8). Also note that another point on the line is (3,4). Now use these two points to find the slope Moving from point (0,8) to (3,4), we go down 4 and to the right 3. Therefore, the slope is - . At this point, we can just substitute into slope intercept form of a line: y = mx + b or

8. First find the slope of the line containing the points (-2,-3) and (2,5): The slope is Now substitute that slope and the point (-4,5) into point slope form of the a line. You get:

y - 5 = 2(x + 4)

EX = RT ln [X]o

zF[X]i

EX = (1.987 cal/deg.mol)(293 deg) ln [X]o

z(23,062 cal/volt.mol)[X]i

OR

EX = (8.315 joules/deg.mol)(293 deg) ln [X]o

z(96,485 joules/volt.mol)[X]i

EK+ = 0.025 ln(12/400) = -0.088 V = -88 mV

ENa+ = 0.025 ln(450/55) = 0.053 V = 53 mV

ECa+2 = 0.0126 ln(10/0.0001) = 0.145 V = 145 mV

ECl- = -0.025 ln(550/56) = -0.058 V = -58 mV

Step-by-step explanation: