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barxatty [35]
3 years ago
10

How many molecules of hydrogen gas are present in a 750 ml container at STP?

Chemistry
1 answer:
BabaBlast [244]3 years ago
4 0

Answer: 1.99 x 10²² molecules H2

Explanation:First we will solve for the moles of H2 using Ideal gas law PV= nRT then derive for moles ( n ).

At STP, pressure is equal to 1 atm and Temperature is 273 K.

Convert volume in mL to L:

750 mL x 1 L / 1000 mL

= 0.75 mL

n = PV/ RT

= 1 atm ( 0.75 L ) / 0.0821 L.atm/ mole.K ( 273 K)

= 3.3x10-² moles H2

Convert moles of H2 to atoms using Avogadro's Number.

3.3x10-² moles H2/ 6.022x10²³ atoms H2 / 1 mole H2

= 1.99x10²² atoms H2

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What observation did Rutherford make from his gold-foil experiment which
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A piece of gold foil was hit with alpha particles, which have a positive charge. Most alpha particles went right through. This showed that the gold atoms were mostly empty space. Some particles had their paths bent at large angles. A few even bounced backward. The only way this would happen was if the atom had a small, heavy region of positive charge inside it.

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3 years ago
At 9°C a gas has a volume of 6.17 L. What is its volume when the gas is at standard temperature?
Alex17521 [72]

Answer:

V₂ = 5.97 L

Explanation:

Given data:

Initial temperature = 9°C (9+273 = 282 K)

Initial volume of gas  = 6.17 L

Final volume of gas = ?

Final temperature = standard = 273 K

Solution:

Formula:

The Charles Law will be apply to solve the given problem.

According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 6.17 L ×  273K /  282  k

V₂ = 1684.41 L.K / 282 K

V₂ = 5.97 L

5 0
3 years ago
Fruit juice when boiled taste sweeter than sucrose because ​
irina1246 [14]

Answer :see explanation

Explanation:

the sweet carbohydrate in fruit is not sucrose , it is fructose.

and fructose tastes sweeter than sucrose

fructose and glucose minus water equals sucrose

and fructose is sweeter than glucose

6 0
2 years ago
What is the de Broglie wavelength, in cm, of a 11.0-g hummingbird flying at 1.20 x 10^2 mph?
KonstantinChe [14]

Answer:1.123 x 10^-31cm

Explanation:

mass of humming bird=  11.0g

speed= 1.20x10^2mph

but I mile = 1.6m

1km=1000

I mile = 1.6x10^3m

1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2

=1.932 x10^5m

recall that  

1 hr= 60 min

1 min=60 secs, 1hr=3600s

Speed = distance/ time

=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s

m= a 11.0g= 11.0 x 10^-3kg

h=6.626*10^-34 (kg*m^2)/s

Wavelength = h/mu

= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)

6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m

but 1m = 100cm

1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm

de broglie wavelength of humming bird = 1.123 x 10 ^-31cm

5 0
3 years ago
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