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3 years ago

8. Why do we see water droplets on the outer surface of a glass containing ice cold water?

Chemistry

2 answers:

baherus [9]3 years ago

5 0

Answer:

8. We see water droplets on the outer surface of a glass containing ice cold water because condensation is happening! This happens when warm water vapour from the surroundings come into contact with a cool surface (in this case the cool glass) and loses heat and condenses, forming water droplets on the surface of the glass.

9 a) -248.15 celcius

b) 99.85 celcius

10. by using 2 syringes, one filled with water and one filled with air. when you compress it, you will find that the one with water barely compresses and the one filled with air should be able to be compressed quite a bit. and to the extent that the syringe is unable to be pushed down further, that is the maximum compressibility of air.

guapka [62]3 years ago

4 0

Answer:

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Explanation:

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You might be interested in

If you have 8.5 moles of Al2(SO4)3 how many grams of the compound do you have?

Zepler [3.9K]

Answer:

$RFM \: of \: aluminium \: sulphate = (2 \times 27) + (32 + (16 \times 4)) \times 3 \\ = 294 \\ 1 \: mole \: weighs \: 294 \: g \\ 8.5 \: moles \: weigh \: (8.5 \times 294) \: g \\ = 2499 \: g$

5 0

3 years ago

When 16.3 g of magnesium and 4.52 g of oxygen gas react, how many grams of magnesium oxide will be formed? Identify the limiting

Tanzania [10]

Answer:

22.77 g.

he limiting reactant is O₂, and the excess reactant is Mg.

Explanation:

From the balanced reaction:

Mg + 1/2O₂ → MgO,

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:

no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.

no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.

So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.

∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).

Using cross multiplication:

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.565 mol of Mg produce → 0.565 mol of MgO.

∴ The amount of MgO produced = no. of moles x molar mass = (0.565 mol)(40.3 g/mol) = 22.77 g.

8 0

3 years ago

Read 2 more answers

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$

Thus, by combining them, we obtain:

$-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}$

Which is related to the general line equation:

$y=mx+b$

Whereas:

$y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}$

It means that we answer to the blanks as follows:

Regards!

8 0

3 years ago

How many grams of calcium chloride are required to react completely with 34.8g of silver nitrate

Mumz [18]

The amount of CaCo3 remaining is 11.5g

3 0

2 years ago

Predict the spontaneity of a reaction (and the temperature dependence of the spontaneity) for each possible combination of signs

Akimi4 [234]

Answer and Explanation:

At constant pressure and constant temperature, the Gibbs free energy of a process is given by the following equation:

ΔG= ΔH - T ΔS

The process is spontaneous when ΔG<0. For this, there are four alternatives depending on the signs of ΔH and ΔS and the temperature (T):

1) ΔH negative, ΔS positive ⇒ b)The reaction will be spontaneous at all temperatures.

ΔG= (-H) - T (+S) ⇒ ΔG<0 always

2) ΔH positive, ΔS negative ⇒ c)The reaction will be nonspontaneous at all temperatures.

ΔG= (+H) - T (-S) ⇒ ΔG>0 always

3) ΔH negative, ΔS negative ⇒ d)The reaction will be spontaneous at low temperature, but nonspontaneous at high temperature.

ΔG= (-H) - T (-S) ⇒ ΔG<0 if TΔS is lower than ΔH, because is the positive term

4)ΔH positive, ΔS positive ⇒ a)The reaction will be nonspontaneous at low temperature, but spontaneous at high temperature.

ΔG= (+H) - T (+S) ⇒ ΔG<0 if TΔS higher than ΔH because is the negative term

8 0

3 years ago