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musickatia [10]
3 years ago
9

What happens when a candle stops burning after putting a glass jar over them

Chemistry
1 answer:
stira [4]3 years ago
8 0

Answer:

The candle ran out of oxygen

Explanation:

There is a concept known as the fire triangle, the three things present for a fire to start and burn. Heat, Oxygen, Fuel. Take any one of the three away and the fire goes out. Pour water on a fire, and it absorbs the heat, put a jar over it, it removes the oxygen, stop adding wood to a campfire it goes out.

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Help me please thank you
meriva
It would be -1 charge
5 0
4 years ago
An electron in the n=6 level of the hydrogen atom relaxes to a lower energy level, emitting light of ?=93.8nm. find the principa
Rama09 [41]

We can find the principal level or lower level using Rydberg's formula:

1/w = R(1/L² - 1/U²) 

<span>where:</span>

<span>w is the wavelength (93.8 nm),</span>

L is the lower energy level (unknown)

U the upper energy level (n= 6)

R is Rydberg's constant (10,967,758 waves per meter)

Substituing known values into the equation:<span>
1/(9.38 * 10^-8 m.) = 10,967,758(1/L² - 1-36) </span>

Using the solver function of the calculator to get for L:

L = 0.999 <span>
so L = 1. 
<span>The lower level is 1 (the ground state).</span></span>

3 0
3 years ago
Read 2 more answers
The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan
aliina [53]

Answer:

K^{2000K}=0.774\\\\K^{3000K}=12.56

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}

Thus, at 2000 K:

\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol

And at 3000 K:

\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

K=exp(-\frac{\Delta _rG}{RT} )

Thus, at each temperature we obtain:

K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

4 0
3 years ago
How do i pee correctly?
Volgvan

Answer:

The perfect pee is by adopting a posture where you sit on the toilet, with you feet flat on the ground, elbows on your knees and you lean forward. This is especially important in children because one in nine children develop bowel and bladder dysfunction purely due to inappropriate posture on the toilet.

Explanation:

6 0
2 years ago
Read 2 more answers
What is the ph of a solution of 0.50 m acetic acid?
frosja888 [35]
You need to use the Ka for the acetic acid and the equilibrium equation.

Ka = 1.85 * 10^ -5

Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)

Ka = [CH3COO-][H+] / [CH3COOH]

Molar concentrations at equilibrium

CH3COOH         CH3COO-     H+

 0.50  - x                  x                 x

Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)

Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50

=> Ka ≈ x^2 / 0.50

=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6

=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030

pH = - log [H+] = - log (x) = - log (0.0030) = 2.5

Answer: 2.5
6 0
3 years ago
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