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Sidana [21]
2 years ago
9

A jar contains helium gas. what would happen if the lid of this Jar was removed

Chemistry
1 answer:
KatRina [158]2 years ago
7 0
The helium gas would simply float out of the bottle and likely spread our.This is because helium is lighter than air and that's why when balloons are filled with them they float.
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Consider a wooden chair and a balloon. What do these two
ICE Princess25 [194]

Answer:

D)

Explanation:

It is the most realistic

6 0
3 years ago
Describe why the periodic table needed to be reordered in 1913.
Norma-Jean [14]

Answer:

1913. Henry Moseley determined the atomic number of each of the known elements. He realized that, if the elements were arranged in order of increasing atomic number rather than atomic weight, they gave a better fit within the 'periodic table'.

Explanation:

GOO

7 0
2 years ago
Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.
dexar [7]

8.8 × 10-5 M is the  [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

 HClO +H2O ⇒ ClO- + H3O+

I  0.265                0        0

C  -x                    +x     +x

E  0.265-x          +x      +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = \frac{[ClO-][H3O+]}{[HClO]}

2.9 × 10^-8 = \frac{[x] [x]}{[0.265-x]}

x^{2} = 7.698 x10^{-9}

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M  in 0.265 M solution of HClO.

8 0
2 years ago
Can sombody pls help me with this​
luda_lava [24]
Wouldn’t it be half of each? For 36 I guess is 18 and 54 will be 27, (NOT SURE)
6 0
2 years ago
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
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