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sertanlavr [38]
3 years ago
11

Ricardo compra un gaseosa en el súper al destaparla la gaseosa comienza a salir espuma y vuelca la mitad al piso ¿Por qué puede

ocurrir eso?
Chemistry
1 answer:
dimaraw [331]3 years ago
8 0

Answer:

Las bebidas gaseosas como las gaseosas están hechas de un soluto de dióxido de carbono gaseoso en un líquido. La solubilidad del dióxido de carbono en el líquido depende de la presión y la temperatura de la lata de refresco, y también de agitar la lata de refresco que introduce burbujas que permanecen ocultas hasta que se abre la lata antes de que burbujee.

Por lo tanto, dado que la presión en la lata de refresco permanece constante, elevar la temperatura, agitar la lata de refresco o congelar el refresco, lo que aumenta la cantidad de dióxido de carbono en la porción líquida, hará que el refresco forme espuma y se derrame.

Explanation:

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1.For the reaction P4 O10(s) + 6H2O(l) → 4H3PO4(aq), what mass of P
Alex17521 [72]
P₄O₁₀ + 6H₂O → 4H₃PO₄
The equation shows us that the molar ratio of
P₄O₁₀ : 6H₂O = 1:6

We also know that one mole of a substance contains 6.02 x 10²³ particles. We can use this to calculate the moles of water.
moles(H₂O) = (5.51 x 10²³) / (6.02 x 10²³)
= 0.92 mole
That means moles of P₄O₁₀ = 0.92 / 6
= 0.15

Each mole of P₄O₁₀ contains 4 moles of P. 
moles(P) = 4 x 0.15 = 0.6 mol
Mr of P = 207 grams per mol
Mass of P = 207 x 0.6
= 124.2 grams
5 0
3 years ago
Read 2 more answers
Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t
Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

3 0
3 years ago
Why polystyrene containers are used to store ice cubes?​
ASHA 777 [7]

Answer:

Hope this may help you

6 0
3 years ago
How much is the oxygen valency?
Marina CMI [18]
The oxygen valency is 2
4 0
2 years ago
Calculate the mass percent of 0.485 g of H, which reacts with oxygen to form 2.32 g H2O
Nataly [62]

Answer:

Mass % of hydrogen = 20.9 %

Explanation:

Given data:

Mass of hydrogen = 0.485 g

Mass of water = 2.32 g

Mass percent of hydrogen = ?

Solution:

Mass % of hydrogen = mass of hydrogen / mass of water × 100

Mass % of hydrogen = 0.485 g/ 2.32 g × 100

Mass % of hydrogen = 0.209  × 100

Mass % of hydrogen = 20.9 %

4 0
3 years ago
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