All categories

3 years ago

What would happen to a gummy bear after soaking for 2 days in distilled water

Chemistry

2 answers:

melomori [17]3 years ago

5 0

Answer:

it would expand

Explanation:

the gummy bear would absorb alot of water

CaHeK987 [17]3 years ago

5 0

Answer:If a gummy bear is soaked in distilled (pure) water for 24 hours, then the gummy bear will increase in size

Explanation:

You might be interested in

Which is a cause of polarity in water molecules? partial negative charge on the hydrogen atoms partial positive charge on the ox

Tamiku [17]

Explanation:

Water is a polar solvent as the hydrogen and oxygen atom has large difference in their electronegativities.

Oxygen atom is highly electronegative as compared to hydrogen atom therefore, it pulls the electrons of hydrogen atom closer towards itself.

As a result, two poles will create forming a partial positive charge on the hydrogen atom and partial negative charge on the oxygen atom.

Thus, we can conclude that high electronegativity difference between oxygen and hydrogen is the cause of polarity in water molecules.

7 0

3 years ago

Why does the amount of energy available change as you move from one trophic level to the next?

olga_2 [115]

Energy decreases as it moves up trophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level. Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels.

6 0

3 years ago

Read 2 more answers

A chemist fills a reaction vessel with 3.82 atm methanol (CH,OH) gas, 7.56 am oxygen (O2) gas, 5.29 atm carbon dioxide (CO2) gas

alisha [4.7K]

<u>Answer:</u> The Gibbs free energy of the given reaction is $1.379\times 10^3kJ$

<u>Explanation:</u>

The equation used to calculate Gibbs free energy change is of a reaction is:

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

For the given chemical reaction:

$2CH_3OH(g)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(CO_2(g))})+(4\times \Delta G^o_f_{(H_2O(g))})]-[(2\times \Delta G^o_f_{(CH_3OH(g))})+(3\times \Delta G^o_f_{(O_2(g))})]$

We are given:

$\Delta G^o_f_{(H_2O(g))}=-228.57kJ/mol\\\Delta G^o_f_{(CO_2(g))}=-394.36kJ/mol\\\Delta G^o_f_{(CH_3OH(g))}=-161.96kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-394.36))+(4\times (-228.57))]-[(2\times (-161.96))+(3\times (0))]\\\\\Delta G^o_{rxn}=-1379.08kJ/mol$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{p}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = -1379.08 kJ/mol = -1379080 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/K mol

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

$Q_{p}$ = Ratio of concentration of products and reactants = $\frac{(p_{CO_2})^2(p_{H_2O})^4}{(p_{CH_3OH})^2(p_{O_2})^3}$

$p_{CO_2}=5.29atm\\p_{H_2O}=3.89atm\\p_{CH_3OH}=3.82atm\\p_{O_2}=7.56atm$

Putting values in above expression, we get:

$\Delta G=-1379080J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(5.29)^2\times (3.89)^4}{(3.82)^2\times (7.56)^3}))\\\\\Delta G=-1379039J=1379.039kJ=1.379\times 10^3kJ$