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Rama09 [41]
3 years ago
7

What were clark's independent and dependent variables

Physics
2 answers:
EleoNora [17]3 years ago
7 0
His full name is dr Kenneth clark

what is being manipulated in the experiment
D.V. (dependent variable): what is being measured
Sloan [31]3 years ago
4 0
I cannot fully answer this question because you have not specifically defined who Clark is. If I knew his full name I might be able to help you more
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The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65  (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0
3 years ago
Read 2 more answers
QUICK giving brainlyest
Soloha48 [4]

The momentum of the second ball was 15 kg.m/s.

<h3>What is inelastic collision?</h3>

In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.

In the given question,

Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.

Initial velocity of 1st ball of mass 3 kg is 1 m/s.

So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.

According to Principle of conservation of linear momentum for this inelastic collision,

Initial momentum of first ball + initial momentum of second ball =  final momentum of the system

⇒   initial momentum of second ball =   final momentum of the system - Initial momentum of first ball

= 18 kg.m/s - 3 kg.m/s.

= 15 kg.m/s.

Hence, initial momentum of second ball = 15 kg.m/s.

Learn more about momentum here:

brainly.com/question/24030570

#SPJ2

5 0
1 year ago
Read 2 more answers
Pls help I’m in a middle of a test and I can’t find the answer
EastWind [94]

Answer:

I believe the answer is B.

5 0
1 year ago
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Confidence intervals for the population mean μ and population proportion p _____ as the size of the sample increases.
Masja [62]

Answer:

becomes narrower

Explanation:

Confidence intervals for the population mean μ and population proportion p becomes narrower as the size of the sample increases.

As,the sample size increases,standard error decreases,so margin of error decreases and hence width of CI decreases.

6 0
3 years ago
What kind of image is formed by an object that is placed beyond the center of curvature on the principal axis of a concave mirro
Delicious77 [7]
As it is outside the focal point it must be real.Real images must be inverted.
As it is beyond the centre of curvature it is also beyond 2F which means that the image is inside the centre of curvature ( between F and 2F from the mirror ) As the image is closer to the mirror than the object it must be diminished in size.

Hope this helps you :)
8 0
3 years ago
Read 2 more answers
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