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3 years ago

What property of argon is uses in light bulbs?

2 answers:

8 0

The filament of the light bulb will get very hot. This will encourage a chemical reaction with most gases that are surrounding that filament - and the result is that the filament burns out. If the filament is in air, it combines with the carbon of carbon dioxide in the air, and the filament disintegrates. But argon is an inert gas - almost nothing reacts with it. So the filament takes a very long time (theoretically infinity) to burn out. But the bulb cannot contain 100% argon: 99.9% is typical; the remaining 0.1% being air. The bulb manufacturers can control the 'life' of a bulb, based on that principle: they do not want their bulbs to last forever!

a_sh-v [17]3 years ago

8 0

it does not react is the answer

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Tpy6a [65]

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3 years ago

How would a solution that is labeled 5.0 M would be read as?

Ivanshal [37]

5 moles because molarity signifies number of moles dissolved in One litre of water

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4 years ago

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Provide a title short explanation or description accompanying an illustration or a photograph

alexdok [17]

A small, green frog wearing a strawberry on its head as a hat

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3 years ago

. Sulfur dioxide can be produced in the laboratory by the reaction of hydrochloric acid and a sulfite salt such as sodium sulfit

shutvik [7]

Answer:

Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g

Explanation:

SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.

the balanced chemical equation is as follows

Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O

Moles of Na₂SO₃ = $\frac{Mass}{Molecular mass} =\frac{25}{126} = 0.198$

Moles of HCl = $\frac{mass}{molecular mass}=\frac{22}{36.5}= 0.6$

using mole ratio method to find limiting reagent

For sodium sulfite $\frac{mole}{stoichiometry} = \frac{0.198}{1}= 0.198$

for HCl $\frac{mole}{stoichiometry} = \frac{0.6}{2}= 0.3$

since sodium sulfite is limiting reactant for above chemical reaction

1 mole of Na₂SO₃ produce 1 mole of SO₂

0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂

∴ Mass of SO₂ produce = mole x molar mass of SO₂

= 0.198 x 64

= 12.672 g

8 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago