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scZoUnD [109]
4 years ago
14

What happened when citric acid and and bicarbonate soda mixed together

Chemistry
1 answer:
nadya68 [22]4 years ago
7 0
The reaction follows the normal rules for an acid-carbonate reaction, which in turn is a special kind of acid-base reaction.

Acid + Carbonate -> Salt + Water + CO2

(Citric Acid) + (bicarbonate soda) -> (<span>Trisodium citrate) + water + CO2</span>

This is a very fast reaction, which produces a lot of carbon dixoide in a very short amount of time, which is the bubbles that you see. The other chemicals produce are clear liquids, so you don't see any of them.
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Looking across period 4 of the periodic table, potassium (atomic number 19) is followed by calcium (atomic number 20), which is
Mumz [18]
While staying in the same period, if we move from left to right across the period, the atomic radius decreases. The reason is, in a period the number of shells remain the same and the number of electrons and protons increase as we move across the period to the right. The increased electrons and protons attract each other with greater force and hence the atomic size decreases. 

So the element on the left most will have the largest atomic radius. So the correct ans is Potassium. Potassium will have the largest atomic size among Potassium, Calcium and Scandium. 
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3 years ago
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Heat makes it softer

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3 years ago
1. write the formula for the following ionic compound: Calcium Carbonate
Salsk061 [2.6K]

Answer:

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2 years ago
Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The
qaws [65]

Answer:

E_{ionization}=5.45\times 10^{-19}\ J

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E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

So, n_i=2 and n_f=\infty (As the hydrogen has to ionize)

Thus,

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J

\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J

E_{ionization}=5.45\times 10^{-19}\ J

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3 years ago
How many significant figures does 0.0006510 of have?
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It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6
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