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2 years ago

A liquid that occupies a volume of 9.0 L has a mass of 7.2 kg. What is the density of the liquid in kg/L

Chemistry

1 answer:

babunello [35]2 years ago

3 0

Answer:

<h2>Density = 0.8 kg/L</h2>

Explanation:

The density of a substance can be found by using the formula

<h3> $Density = \frac{mass}{volume}$ </h3>

From the question

mass = 7.2 kg

volume = 9.0 L

Substitute the values into the above formula and solve for the density

That's

<h3> $Density = \frac{7.2}{9}$ </h3>

We have the final answer as

<h3>Density = 0.8 kg/L</h3>

Hope this helps you

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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

Determine the number of moles in 4.21 x 10^23 molecules of CaCI2?????

Alla [95]

Explanation:

Given parameters:

Number of molecules = 4.21 x 10²³ molecules

Unknown parameters:

Number of moles

Solution:

A mole can be defined as the amount of a substance that contains the avogadro's number of particles i.e 6.02 x 10²³

To find the number of moles:

Number of moles = $\frac{number of molecules }{[tex]6.02 x 10^{23}$ }[/tex]

Number of moles = $\frac{4.21 x 10[tex]^{23}$ }{ $6.02 x 10^{23}$ }[/tex]

Number of moles of CaCl₂ = 0.699moles

Learn more;

mole calculation brainly.com/question/13064292

#learnwithBrainly

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3 years ago

Why might a major volcanic eruption lead to cooler temperatures over a large area around the volcano ?

Readme [11.4K]

The dense aerosol cloud created caused decreases in the amount of solar radiation that reached the Earths surface.

5 0

3 years ago

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

8 0

3 years ago

Read 2 more answers

Sample of 200mls of 0.5 sulphuric acid,was asked to produce 1.2M of the new solution.Calculate the volume of the new solution

AlladinOne [14]

Answer:

$V_2=83.3mL$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of the new solution by using the general formula of dilution:

$V_1M_1=V_2M_2$

In such a way, we solve for the final volume, V2, to obtain:

$V_2=\frac{200mL*0.5M}{1.2M}\\\\V_2=83.3mL$

Regards!

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3 years ago