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2 years ago

Why does sugar sink to the bottom of a tea cup?

Chemistry

2 answers:

Mandarinka [93]2 years ago

4 0

Because it is more dense than the water or tea.

miskamm [114]2 years ago

4 0

Answer:

The tea (solution) is hot, this means that the solution molecules are moving in a fast rate. This results in gaps being present between the solution molecules, when the sugar (solute) is added its molecules can fit in between the gaps making its way down the containers with ease.
Solubility is directly proportional to temperature, thus an increase in temperature (of the solution) will result in an increase in solubility (of the solute).

Explanation:

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Enter the correct ground-state (or lowest energy) configuration based on the number of electrons: 1s^4 2s^4 2p^12.

Blababa [14]

Answer:

The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.

or the

FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.

Element Atomic number Electron configuration

sulfur 16 1s22s22p63s23p4

chlorine 17 1s22s22p63s23p5

argon 18 1s22s22p63s23p6

or the

Two electrons

Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.

Explanation:

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6 0

2 years ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][Br^-]$

$4.2\times 10^{-8}=0.073\times [Br^-]$

$[Br^-]=5.75\times 10^{-7}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgBr\rightleftharpoons Ag^++Br^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][Br^-]$

$7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M$

$[Ag^+]=1.34\times 10^{-6}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{1.34\times 10^{-6}}{0.073}\times 100$

Percent of $Ag^+$ remains = 0.0018 %

3 0

3 years ago

Which substances in the diagram are most unstable? positive values of enthalpy negative values of enthalpy negative values -50 t

NemiM [27]

Answer:

no s e

Explanation:

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8 0

3 years ago

Read 2 more answers

List the spectator ions in the following reaction.

Rudik [331]

Answer:

2Na⁺ (aq) and 2OH⁻(aq)

Explanation:

Spectator ions:

Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.

Given ionic equation:

Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)

In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq) net ionic equation can be written as,

Net ionic equation:

Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)

4 0

2 years ago

The most reactive metals are the ________ to corrode.

katrin [286]

Answer:

2. quickest

Explanation:

because they have lower activation energy.

3 0

3 years ago