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xxMikexx [17]
3 years ago
9

Why do teapots have highly polished surface

Chemistry
2 answers:
Natali [406]3 years ago
6 0

Good conductors of electricity and heat, form cations by loss of electrons,

LekaFEV [45]3 years ago
6 0
They are good conductors of heat
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Identify the number with the correct number of significant figures for each step of the calculation and for the final answer. No
lara [203]

Answer: The answer is 167

Explanation: This is because that was right on edg. so yea heart this tho plsss

7 0
3 years ago
Read 2 more answers
What is a mole?
leva [86]
It's hard to relate a mole to carbon or sulfur. Imagine if I walked up to you and said, "What's the relation between a dozen and donuts?"

A mole is a form of measurement for atoms, more specifically, 6.02 * 10^23 atoms. I suppose you could relate it to Carbon or Sulfur, since the number of atoms of each are usually measured in moles.

Carbon and Sulfur don't have a set number of moles (Just like donuts don't have to be a dozen), so it's hard to answer your second question.
In the atomic table, the number you see under the element is the molar mass, which is the weight of an a mole of the element. In this way, I guess there's a mole of Carbon and Sulfur present, if we're looking at the periodic table.

-T.B.
8 0
3 years ago
The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−
barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

Expression for an equilibrium constant K_c:

K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]

Solubility product of the reaction:

K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

Reaction between Gibb's free energy and equilibrium constant if given as:

\Delta G^o=-2.303\times R\times T\times \log K_c

\Delta G^o=-2.303\times R\times T\times \log K_{sp}

\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]

\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

Gibb's free energy when concentration [Ag^+] = 1.0\times 10^{-2} M and [Br^-] = 1.0\times 10^{-3} M

Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
  • For reaction to non spontaneous reaction:  \Delta G>0.

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
Read 2 more answers
The density of NaCl( s) is 2.165 g cm 3 at 25 C. How will the solubility of NaCl in water be affected by an increase in pressure
sammy [17]

The solubility of NaCl in water will not be affected by an increase in pressure.

We know that the density of NaCl(s) in 2.165 g/cm³ at 25 °C and we want to know how will its solubility in water be affected when the pressure is increased.

<h3>What is solubility?</h3>

Solubility is the maximum mass of a solute that can be dissolved in 100 grams of solvent at a determined temperature.

The solubility of a solid, such as NaCl, in a liquid, is mainly affected by the temperature. However, since solids are not compressible, an increase in pressure will not affect its solubility.

On the other hand, the solubility of gases in water will increase with an increase in pressure, as stated by Henry's law.

The solubility of NaCl in water will not be affected by an increase in pressure.

Learn more about solubility here: brainly.com/question/11963573

7 0
2 years ago
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