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Oduvanchick [21]
2 years ago
12

For her science project, Paula wants to determine if the growth of bacteria in standing rainwater changes over time. She collect

s rainwater and distributes it evenly in 6 jars and places them all indoors in a spot that does not receive a lot of sunlight to limit evaporation. Using a bacteria testing kit, she measures the bacteria in the water in the first jar on the 5th day, second jar on the 7th day, third jar on the 9th day, fourth jar on the 14th day, fifth jar on the 20th day, and sixth jar on the 24th day. What is the independent or manipulated variable in Paula's experiment?
Chemistry
1 answer:
tamaranim1 [39]2 years ago
8 0

Answer:

The correct answer is - the number of days the rainwater stands in the jars.

Explanation:

The independent variable or the manipulating variable is a variable in a research or study is the variable which affects the dependent variable or the variable that is going to measure in the study.

In the study, variables that recieves the treatment and changed in every subject are independent variables and in this study number of days the rainwater standing in various jars that might affect the growth of the bacteria in rainwater.

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Answer: C

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2 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
Find the number of moles of water that can be formed if you have 126 mol of hydrogen gas and 58 mol of oxygen gas.
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Since a water molecule is H2O, you would divide 126 hydrogen molecules by 2, and you would get 63. That means you have 63 double hydrogen molecules, and 58 oxygen molecules to pair up with them. So that means you could have 58 molecules of water, with 5 double hydrogen molecules, so basically 10 extra molecules of hydrogen along with the H2O molecules. Hope I helped! :)

6 0
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