Due to the increase of acid rain over the last few decades, the pH levels in lakes and streams seem to be decreasing. While certain
animals can tolerate low pH levels, others are extremely sensitive to the effects of acid rain and cannot live under the conditions.
Candice is working on a project to determine whether acid rain is decreasing the amount of bass and mayflies in a lake in her
hometown. She is using the scientific method. To MOST accurately test the pH of the lake, Candice should useas
part of her experimental design procedures.

A) litmus paper
B) hydrion pH papers
C) a portable digital pH meter
D) a naturally occurring indicator like red cabbage juice

A portable digital pH meter will be best suited.

The pH is the negative logarithmic expression of the concentration of hydrogen ions present in a solution. As its negative of the log of hydrogen ion concentration, so less it is, more is the hydrogen ion concentration. So a lower pH is acidic and higher is basic.

Now different indicators are used to measure the pH of a solution. Indicators can measure the pH of a solution both qualitatively and quantitatively.

Among the options given here, the litmus and cabbage juice are only qualitative measurers. So they aren't suitable here. Now the hydrion pH papers have the accuracy from 0.5 pH units to 0.25 pH units. Whereas the pH meter has the accuracy of 0.2 pH units to upto 0.01pH units. So the digital meter is much more accurate and scientific. As Candice is running her experiment most accurately, so she must use digital pH measurers.

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

i belive it is water