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3 years ago

The ___ is the best part of an experiment that is not being tested and it used for comparison

Chemistry

2 answers:

Anettt [7]3 years ago

5 0

The answer is control

aleksandr82 [10.1K]3 years ago

3 0

The control is the part of an experiment that is not being tested and it is used for comparison.
The control is the best way for us to understand how the variables effect our experiments.

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

A solid that forms and separates from a liquid mixture is a.......

inna [77]

A solid that forms and separates from a liquid mixture is a chemical change.

4 0

3 years ago

How many grams of ca are needed to react completely with 2.20 L of a 4.50 m hcl solution

Dmitry_Shevchenko [17]

Ca + 2HCl = CaCl₂ + H₂

c=4.50 mol/l
v=2.20 l

n(HCl)=cv

m(Ca)/M(Ca)=n(HCl)/2

m(Ca)=M(Ca)cv/2

m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g

198 grams of Ca are needed

5 0

3 years ago

Read 2 more answers

Calculate the potential energy of a 25kg ball sitting 25m high

Mariulka [41]

The answer is 6,125. To get this you multiply both by 9.8

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3 years ago

A baker touches a pie right after taking it out of the oven. Which statement best explains why the pie feels hot?

ANTONII [103]

Answer:

Well could you please show the answers?