Ask question

Ask question

All categories

Gelneren [198K]

3 years ago

14

5.6 x 10^3 cm = ___cg

1 answer:

NeX [460]3 years ago

3 0

Answer:

560000 cg

Explanation:

5.6 x 10^3 = 5600

1 g = 100 cg

5600 g = 560000 cg

You might be interested in

A team of researchers are working on a project to make a new kind of airplane fuel. During their experiment, there was an explos

tiny-mole [99]

Well for a start, this makes absolutely no sense, "discovered a fuel that burns so hot that it becomes cold."

<span>And yes, it's not science if the experiment can't be repeated. In fact they should WANT it to be repeated so that you can get credit for discovering something new and then possibly harness this effect to produce useful applications. </span>

<span>For all we know they had a fewer of LN2 in the lab that got shredded by the blast, LN2 could certainly have frozen many things (not metal though, since metal is already solid at room temperature, (except for mercury)), and afterwards would leave no trace.</span>

6 0

3 years ago

What is the ph of 0.1 m naoh solution??

Rudik [331]

NaOH is a strong base so pH will be around 13 to 12. Whatever number of moles of NaOH you have approximately the pH of NaOH will be around 14 13 or 12

7 0

3 years ago

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

Select the letter of a category on which you should focus when proofreading.

oee [108]

B. Who, what, when, where, why, and how. Those are the most important and key componants in a story.

8 0

3 years ago

I need help with this chem study guide

Cloud [144]

Can you please provide a clearer image?
Thanks

7 0

2 years ago