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2 years ago

Which term names elements that are usually shiny and good conductors of heat and electricity?

Chemistry

1 answer:

Sonbull [250]2 years ago

4 0

Elements that are usually shiny and good conductors of heat and electricity are metals.

<h3>What are conductors?</h3>

The conductors are the metal or non metal elements which allow heat and electricity to easily pass through them.

Metals are elements with good conductivity of electric current and heat. Metals have lustre characteristics. They are shiny and bendable just like copper wire, which is a metal.

Hence, elements that are usually shiny and good conductors of heat and electricity are metals.

Learn more about conductors.

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gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

7 0

3 years ago

If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams

Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of $KCl_3$ = 2.0 moles

Molar mass of $O_2$ = 32 g/mole

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 2.0 moles of $KClO_3$ react to give $\frac{2.0}{2}\times 3=3.0$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g$

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}$

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = $214.0^oC=273+214.0=487K$

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

$\rho$ = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

$P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}$

$P=1.78atm$

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0

3 years ago

What is the mass in grams of 85.32 mL of blood plasma with a density of 1.03 g/mL

sveticcg [70]

<span>Well if you're looking for grams, all you need to do is cancel out units. (ml)(g/ml)=g because the ml cancels out. Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :) Good luck! Hope that helped. When in doubt, look at the units.</span>

3 0

3 years ago

As refrigerant passes through the expansion valve, it cools down or heats up?

Yuri [45]

It cools down i am 100 percent sure it cools down

5 0

3 years ago

Read 2 more answers

A process at constant T and P can be described as spontaneous if ΔG < 0 and nonspontaneous if ΔG > 0. Over what range of t

creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

2 SO3(g) → 2 SO2(g) + O2(g)

ΔHf° (kJ mol-1) -395.7 -296.8

S° (J K-1 mol-1) 256.8 248.2 205.1

ΔH° = kJ

S° = J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS = ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) = 2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ = 197.8 - T(0.1879)

and the reaction will become spontaneous when the term T(0.1879) becomes greater that 197.8,

0 = 197.8 - 0.1879 T ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0

3 years ago