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2 years ago

What reaction would take place if a hot tungsten filament bulb was surrounded by air

Chemistry

2 answers:

expeople1 [14]2 years ago

4 0

I don't think it wont be a big explosion

ira [324]2 years ago

3 0

The filament wire will burn away after some time since argon is used in light bulbs as it is a noble gas, having a full outer covalent shell of 8, making it unreactive and stable, thus preventing the filament from burning. Air is a different case....

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A photographic "stop bath" contains 160ml of pure acetic acid, HC2H302(1) in 650ml solution. what is the v/v concentration of ac

Butoxors [25]

Answer:

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um belo dia abençoado

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2 years ago

A toy car is placed at 0 on a number line. It moves 9 cm to the left, then 4 cm to the right, and then 6 cm to the left. What is

snow_lady [41]

Answer:

Explanation:

The total distance traveled by the toy cay would be 19 cm.

The total distance traveled should not be mistaken for total displacement. While displacement measures the distance and direction from the starting position of the toy car relative to its final position, the total distance traveled is calculated by adding all the movements of the toy car together. Hence;

Total distance traveled = 9 + 4 + 6 = 19 cm

7 0

2 years ago

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495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

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Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base

Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

$\frac{r_{cation}}{r_{anion}}$

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, $Na^+$ = 102 pm

Radius of cation, $Br^-$ = 196 pm

$\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520$

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0

2 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 1.50 grams of aluminum foil in a solut

Lisa [10]

Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Explanation: $2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu$

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.

Thus 54 g of of aluminium react with 270 g of copper chloride.

1.50 g of aluminium react with= $\frac{270}{54}\times 1.50=7.5 g$ of copper chloride.

3 moles of copper chloride gives 3 moles of copper.

7.5 g of copper chloride gives 7.5 g of copper.

8 0

2 years ago

Read 2 more answers