Answer:
2Fe(s) + 3O2(g) --------> 2FeO3(s)
Explanation:
According to the question, a battery was used to light the steel wool by bringing the terminals very close together. When the battery came into contact with the steel wool, current was sent out through the thin wire. This caused the iron to heat up quite well.
Iron reacts with oxygen under these conditions as follows;
2Fe(s) + 3O2(g) --------> 2FeO3(s)
This is the chemical reaction that occurs when the steel wool is set on fire.
Bonjeur monseur, la french are here! it seems you have a problem with the elements on earth. lead seems correct since it is indeed a harmful material
Answer:
Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)
Explanation:
Part 1. Volume of reactant
(a) Balanced chemical equation.
![\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl](https://tex.z-dn.net/?f=%5Crm%202Na_%7B3%7DPO_%7B4%7D%20%2B%203CuCl_%7B2%7D%20%5Clongrightarrow%20Cu_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%20%2B%206NaCl)
(b) Moles of CuCl₂
![\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20CuCl%7D_%7B2%7D%20%3D%5Ctext%7B%2016.7%20mL%20CuCl%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B0.200%20mmol%20CCl%7D_%7B2%7D%7D%7B%5Ctext%7B1%20mL%20CuCl%7D_%7B2%7D%7D%20%3D%20%20%5Ctext%7B3.340%20mmol%20CuCl%7D_%7B2%7D)
(c) Moles of Na₃PO₄
The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂
![\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%20%3D%20%20%5Ctext%7B3.340%20mmol%20CuCl%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%7B%5Ctext%7B3%20mmol%20CuCl%7D_%7B2%7D%7D%20%3D%5Ctext%7B2.227%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D)
(d) Volume of Na₃PO₄
![V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}](https://tex.z-dn.net/?f=V%20%3D%20%5Ctext%7B2.227%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mL%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%7B%5Ctext%7B0.300%20mmol%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%20%3D%20%5Ctext%7B7.42%20mL%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%20%5C%5C%5C%5C%5Ctext%7BThe%20reaction%20requires%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B7.42%20mL%20Na%24_%7B3%7D%24PO%7D_%7B4%7D%7D%24%7D)
Part 2. Net ionic equation
(a) Molecular equation
![\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})](https://tex.z-dn.net/?f=%5Crm%202Na_%7B3%7DPO_%7B4%7D%28%5Ctext%7Baq%7D%29%20%2B%203CuCl_%7B2%7D%28%5Ctext%7Baq%7D%29%20%5Clongrightarrow%20Cu_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%28%5Ctext%7Bs%7D%29%20%2B%206NaCl%28%5Ctext%7Baq%7D%29)
(b) Ionic equation
You write molecular formulas for the solids, and you write the soluble ionic substances as ions.
According to the solubility rules, metal phosphates are insoluble.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>
The net ionic equation is
3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)
Answer:I belive it would be attracted seeing as how there are more magmatic charges on that side of no 2 and how there are more positive charges on the middle side of ballon no1.
Explanation:
The heat required to raise the temperature to a specific temperature change of a sample is related to the specific heat capacity of the substance. In this case, the heat can be calculated through mCpΔT = 350 g * 0.39 J/g C *25 C. This is equal to 3412. 5 Joules. Closest answer is C.