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3 years ago

How do you describe how sugar dissolves in tea using the big idea of particles?

2 answers:

4 0

If you use big pieces of sugar, the particles will dissolve harder...As the pieces are smaller, as much they dissolve faster.

Komok [63]3 years ago

3 0

Sugar is an ionic compound and tea is most likely a polar substance. When the sugar is placed in the tea the molecules of the tea attack the sugar molecules using their negative and positive pulls. The negative sides of the tea molecules pull at the positive parts of the sugar, and vice versa. <span />

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2 years ago

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3 years ago

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Which element in Group 17 is the most active nonmetal? (1) Br (2) I (3) Cl (4) F

Maksim231197 [3]

The element of the group 17 that is most active non metal is fluorine.

The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).

Among all the elements of the group 17. Fluorine is the smallest in size.

Because of the small size of fluorine it has the highest electronegativity in group 17.

This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.

Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.

It reacts readily to form oxides and hydroxides.

So, we can conclude here that fluorine is the most active non metal of group 17.

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9 months ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$