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3 years ago

Why LED is better option for testing the conduction of electricity?

Chemistry

1 answer:

Anettt [7]3 years ago

5 0

Answer:

Led is a light-emitting diode, a semiconductor diode that glows when a voltage is applied. So, there is always a sure result of whether a substance conducts electricity or not as the light glows. That is why it is a better option for testing conduction of electricity.

Explanation:

You might be interested in

Name three pure elements and three pure compounds

VARVARA [1.3K]

Answer:

Pure elements:

Hydrogen(H)

Nitrogen(N)

Magnesium (Mg)

Pure Compounds:

Oxygen gas(O2)

Water (H2O)

Ammonia (NH3)

Explanation:

Hope it helps.Please mark me brainliest.^-^

5 0

3 years ago

Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.

Oduvanchick [21]

Answer:

103.9 g

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First we convert 54.0 g of propane (C₃H₈) into moles, using its molar mass:

54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we convert 1.23 moles of C₃H₈ into moles of CO₂, using the stoichiometric coefficients:

1.23 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 3.69 mol CO₂

We convert 3.69 moles of CO₂ into grams, using its molar mass:

3.69 mol CO₂ * 44 g/mol = 162.36 g

And apply the given yield:

162.36 g * 64.0/100 = 103.9 g

7 0

3 years ago

The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti

lisabon 2012 [21]

Answer:

Option E, Half life = $2.96\times 10^3\ s$

Explanation:

For a first order reaction, rate constant and half-life is related as:

$t_{1/2}=\frac{0.693}{k}$

Where,

$t_{1/2}$ = Half life

k = Rate constant

Rate constant given = $2.34\times 10^{-4}\ s^{-1}$

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{2.34 \times 10^{-4}}=2.96\times 10^3\ s$

So, the correct option is option E.

4 0

4 years ago

Which two substances are covalent compounds? * C6H12O6(s) and KI(s) C6H12O6(s) and HCl(g) KI(s) and NaCl(s) NaCl(s) and HCl(g)

Ray Of Light [21]

Answer: Option (b) is the correct answer.

Explanation:

A covalent compound is defined as the compound in which sharing of electrons take place between the combining atoms. Generally, when two or more non-metals chemically combine together the it will lead to the formation of a covalent compound.

For example, $C_{6}H_{12}O_{6}$ and HCl is also a covalent compound.

And, a compound in which transfer of electrons occur between the combining atoms is known as an ionic compound. Whenever, a metal chemically combines with a non-metal then it will always lead to the formation of an ionic compound.

For example, KI is an ionic compound.

Thus, we can conclude that $C_{6}H_{12}O_{6}$ and HCl are the two substances which are covalent compounds.

7 0

3 years ago

djyliett [7]

Answer:

Explanation:

1. The amount of CaCO3 must be so small that

is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full

required for equilibrium.

3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)

[

]

[

]

[

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)

[

]

[

]

[

]

; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

Hope this helps :)

3 0

3 years ago