In this assignment, you and your partners will design a plan for how a city can use a piece of land. You will begin by conducting Internet research on soil erosion. The information you collect should be recorded in the table provided on your Student Guide Worksheet. Then, you will read about a fictional city and how it would like to develop a piece of land. The city wants to use the land wisely so they can maintain the soil and increase biodiversity. Your background research on soil erosion will help you learn why soils are valuable, what human activities increase soil erosion, and what things can be done to sustain soils. With your partners, you will brainstorm a list of possible ways the city can develop this land. You will then evaluate each way based on how it will affect the environment, how much money is needed to develop the land and how the developed land will benefit the citizens of the city. Finally, you will create a presentation summarizing your ideas and present your design plan to the class.

i look it up online and i'll tell you

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.