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Yakvenalex [24]
3 years ago
15

Which of the following would be considered important in including in the summary of an experimental investigation?

Chemistry
2 answers:
just olya [345]3 years ago
7 0

Answer is: variables that were manipulated during the experiment.

The two main variables in an experiment are the independent and dependent variable.

Dependent variable is the variable being tested and measured in a scientific experiment.

Dependent variables depend on the values of independent variables. The dependent variables represent the output or outcome whose variation is being studied.

Laboratory room number is the least important.

saveliy_v [14]3 years ago
5 0

NOOOO GUYS ITS VARIABLES THAT WERE MANIPULATED DURING THE EXPERIMENT i took the test i do homeschool k12

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nordsb [41]

Answer: The entire water/ice solution is at the melting/freezing point, 32°F (0°C). Adding rock salt — or any substance that dissolves in water — disrupts this equilibrium.

Explanation: Hope this helps! Have a great day :)

7 0
3 years ago
What is enthalpy?
aleksandrvk [35]

Answer: the heat content of a system at constant pressure.

Explanation:

Enthalpy is defined as the heat content of a system at constant pressure.

It is the heat absorbed or released during a reaction at constant pressure,denoted as ΔH.

7 0
3 years ago
There is a structure of CH3CH2CHCHCH2CH3, with a –CH2CH3 group attached to the third and fourth carbons.
NARA [144]

Explanation:

2,3-diethyl hexane

At first we select a long chain.

Then, we number that chain from that side where substituent position is closer.

Then, we write it's IUPAC name

Position of substituent + substituent name + chain name + suffix

Here,

2,3 + -diethyl + hex + -ane

= 2,3-diethyl hexane

7 0
2 years ago
A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm
Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys)  =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

<u>Step 1</u>: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

<u>(a) reversibly</u>

<u>Step 2:</u> Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

<u>Step 3:</u> Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

<u>Step 4:</u> Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)  

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

<u>(b) against a constant external pressure of 1.00 atm</u>

<u>Step 1</u>: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

   ⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

     ⇒ Vf = (1*0.08206 *300)/1

      ⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

<u>Step 2:</u> Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

<u>Step 3: </u>Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0
3 years ago
What is the name of the compound CH4?*
solmaris [256]

Answer:

Methane

Explanation:

Methane is a synthetic compound with the substance recipe CH4. It is a gathering 14 hydride and the most straightforward alkane, and is the principle constituent of petroleum gas

6 0
3 years ago
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