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GalinKa [24]
2 years ago
11

Sara wanted to perform a scientific experiment relating the effect of ultraviolet (UV) light on the survival rate of microscopic

protists in pond water. She went to a local pond and collected a large bucket of pond water, then poured 500 milliliters of the pond water into each of two identical glass jars. She viewed a small sample of the pond water under a compound light microscope to verify that the water contained multiple species of microscopic protozoa. She was also able to get a rough count of the organisms in her sample using a special microscope slide. To set up her experiment Sara took one of the glass jars with pond water and protists, labeled it “100”, and placed it 10 centimeters under a lamp that transmits 100% UV light. The other jar, she labeled “O” and placed it 10 centimeters under a lamp that transmits 0% UV light. After the pond water/protist mixture in both jars was exposed to the lights for 8 hours, she collected samples from both jars. She used the same procedure as before and made a 2nd count of the microorganisms to compare to her first count.
12. If Sara were to graph her data using a bar graph, what should her axes be labeled?
​
Chemistry
1 answer:
posledela2 years ago
7 0
Ummmm hold up one sec
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When 0.270 mol of a nondissociating solute is dissolved in 410.0 mL of CS2, the solution boils at 47.52 ∘C. What is the molal bo
grandymaker [24]

Answer:

Kb = 0.428 m/°C

Explanation:

To solve this problem we need to use the <em>boiling-point elevation formula</em>:

  • <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.

So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.

molality = mol solute / kg solvent

  • Density of CS₂ = 1.26 g/cm³
  • Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg

molality = 0.270 mol / 0.5166 kg = 0.5226 m

Now we <u>solve for Kb</u>:

<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m

  • 47.52 °C - 46.3 °C = Kb * 0.5226 m
  • Kb = 0.428 m/°C
3 0
2 years ago
If you gently shake a carbon dioxide fire extinguisher, you will feel the presence of liquid within the extinguisher What condit
scoundrel [369]

When we wish to convert a gas to liquid we have to either

a) decrease temperature

b) increase pressure

In case of fire extinguisher the CO2 is found to be in liquid state, this is as the CO2 is pressurized at high pressure which keeps CO2 in liquid state

the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas

Pressure = 5 - 73 atm

Temperature = -57  to 31 degree Celsius


6 0
2 years ago
Calculate the number of kilojoules of energy required to convert 50.0 grams of solid DMSO initially at a temperature of 19.0°C t
GuDViN [60]

Answer:

20.79 kilojoules

Explanation:

Using Q = m×c×∆T

Where;

Q = Quantity of heat (J)

c = specific heat capacity of solid DMSO (1.80 J/g°C)

m = mass of DMSO

∆T = change in temperature

According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C

Q = m×c×∆T

Q = 50 × 1.80 × (250°C - 19°C)

Q = 90 × 231

Q = 20790 Joules

To convert Joules to kilojoules, we divide by 1000 i.e.

20790/1000

= 20.79 kilojoules

Hence, 20.79 kilojoules of energy is required to convert 50.0 grams of solid DMSO to gas.

4 0
2 years ago
Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
Margarita [4]

Answer:

thickness = 0.29 cm

Explanation:

In order to make fake iron ball [made of gold] we should get mass of fake ball

should be equal to that of Iron ball.So for that we should calculate

volume of iron ball using diameter given ;formula is 4/3 pi r^3

given d= 6 cm;so radius r= 6/2 = 3 cm

then volume of Iron ball = 4/3 *3.14* 3^3 = 113.04 cm^3

So mass of iron ball = volume x density = 113.04 * 5.15 g/cm^3 = 582.156 g

This must be the mass of gold ball ;now calculate volume of gold ball using its

density

volume of gold ball = mass of gold ball/density of gold ball = 582.156 g/19.3 g/cm^3

= 30.1635 cm^3

Now this must be the volume of hollow sphere whose outer radius R = 3cm

and inner radius r= ??

Volume of hollow ball = 4/3pi[R3-r^3]

30.1635 cm^3 = 4/3 pi [3^3-r^3]

30.1635* 3/4*3.14 = 27 - r^3

7.2046 = 27- r^3

r^3 = 19.7954

r= 2.7051 cm

So the thick ness = outer radius- inner radius = 3 - 2.7051 = 0.2949 cm

rounding to 2 significant figures

we get thickness = 0.29 cm

8 0
2 years ago
A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 417 pm. The diameter of the metal atom is:
enyata [817]

Answer:

b. 295 pm

Explanation:

To answer this question we need to use the equation of a face-centered cubic laticce:

Edge length = √8 R

<em>Where R is radius of the atom.</em>

<em />

Replacing:

417pm = √8 R

R = 147.4pm is the radius of the atom

As diameter = 2 radius.

Diameter of the metal atom is:

147.4pm* 2 =

295pm

Right answer is:

<h3>b. 295 pm </h3>

8 0
2 years ago
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