Answer: 2) 2HCl(sq) + CaCO3(s) CaCl2(sq) + CO2(g) + H2O (l) No of moles of CaCO3 = amount of the CaCO3 (g)/mw of CaCO3 (g/mole)= 0.8085 g/100 g/mole = 0.008085
Explanation:
Answer:
This process is known as doping. It can be done by adding either of two types of impurity to the crystal.
(A) By adding electron rich impurities i.e., group 15 elements to the silicon and germanium of group 14 elements.
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Answer:

Explanation:
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In this case, since the reaction for the formation of ammonia is:

We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:

And the result is:

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Answer:
<em>The correct option is A) The atom has changed over the years.</em>
Explanation:
Option A is not correct because an atom does not change. It remains the same always. The structure of an atom will always contain neutrons, protons and electrons.
Scientific theories tend to change over time as new information is brought up. Novel techniques lead to more advanced experiments and results. As a result, a scientific theory is subjected to change whenever evidence is found for a new theory which challenges the older one.
The atomic theory has changed many times due to novel scientific techniques and experiments.
Answer:
2.387 mol/L
Explanation:
The reaction that takes place is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
First we <u>calculate how many moles of each reagent were added</u>:
- HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
- Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂
460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.
Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:
- 203.85 mmol HCl *
= 101.925 mmol Ba(OH)₂
This means the remaining Ba(OH)₂ is:
- 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂
There are two OH⁻ moles per Ba(OH)₂ mol:
- OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻
Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):
- 716.15 mmol OH⁻ / 300.0 mL = 2.387 M
So the answer is 2.387 mol/L