An ion of chlorine element has a larger radius than an atom of the same element. It has a negative charge.
Answer:
The relevant equation is:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Explanation:
1 mol of calcium carbonate can react to 2 moles of Hydrochloric acid to produce 1 mol of water, 1 mol of calcium chloride and 1 mol of carbon dioxide.
The formed CO₂ is the reason why you noticed bubbles as the reaction took place
I googled it and what I found was: "Vaporization is a transitional phase of an element or compound from a solid phase or liquid phase to a gas phase while evaporation is a type of vaporization wherein the transition from a liquid phase to a gas phase takes place below the boiling temperature at a given pressure, and it occurs on the surface."
But my science teacher always told us they were both pretty much the same thing: liquid to gas. There isn't a major difference and if you see one of the words on a test just remember - liquid to gas!
Answer:
2.05mg Fe/ g sample
Explanation:
In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.
In the problem you recover 3.5mg Fe / 1.0101g sample: <em>3.465mg Fe / g sample. </em>As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:
3.465 / 4×100 = <em>86,6%</em>
As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:
1.7mg Fe / 0.9582g sample× (100/86.6) = <em>2.05mg Fe / g sample</em>
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I hope it helps!
Answer:
,
, ![x_{3} = 1.384](https://tex.z-dn.net/?f=x_%7B3%7D%20%3D%201.384)
Explanation:
The expression for the approximation via Newton's Method has the following form:
![x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}](https://tex.z-dn.net/?f=x_%7Bn%2B1%7D%20%3D%20x_%7Bn%7D%20-%20%5Cfrac%7Bf%28x_%7Bn%7D%29%7D%7Bf%27%28x_%7Bn%7D%29%7D)
The function and its derivative are, respectively:
![f(x_{n}) = x^{5}-5](https://tex.z-dn.net/?f=f%28x_%7Bn%7D%29%20%3D%20x%5E%7B5%7D-5)
![f'(x_{n})= 5\cdot x ^{4}](https://tex.z-dn.net/?f=f%27%28x_%7Bn%7D%29%3D%205%5Ccdot%20x%20%5E%7B4%7D)
After substituting the known variable, the Newton's expression is left as follows:
![x_{n+1} =x_{n}- \frac{x_{n}^{5}-5}{5\cdot x_{n}^{4}}](https://tex.z-dn.net/?f=x_%7Bn%2B1%7D%20%3Dx_%7Bn%7D-%20%5Cfrac%7Bx_%7Bn%7D%5E%7B5%7D-5%7D%7B5%5Ccdot%20x_%7Bn%7D%5E%7B4%7D%7D)
The first two iterations are presented herein:
![x_{2} = 1.6 - \frac{1.6^{5}-5}{5\cdot (1.6)^{4}}](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%201.6%20-%20%5Cfrac%7B1.6%5E%7B5%7D-5%7D%7B5%5Ccdot%20%281.6%29%5E%7B4%7D%7D)
![x_{2} = 1.433](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%201.433)
![x_{3} = 1.433 - \frac{1.433^{5}-5}{5\cdot (1.433)^{4}}](https://tex.z-dn.net/?f=x_%7B3%7D%20%3D%201.433%20-%20%5Cfrac%7B1.433%5E%7B5%7D-5%7D%7B5%5Ccdot%20%281.433%29%5E%7B4%7D%7D)
![x_{3} = 1.384](https://tex.z-dn.net/?f=x_%7B3%7D%20%3D%201.384)