Answer:
The law of conservation of mass states that mass is neither created nor destroyed but the mass of the system must remain constant over time. The total number of atoms in the reactants is equal to the total number of atoms in the product. Therefore, this chemical equation shows that energy is conserved and demonstrates the law of conservation of mass.
Answer:
Neutral ions
Explanation:
Because they have a neutral charge They can only produce neutrally.
Answer:
- Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.
Explanation:
<u>1. Adding AgNO₃ to NaCl solution:</u>
- AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
<u>2. Adding AgNO₃ to NaClO₃ solution</u>
- AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)
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<u>3. Relevant solubility rules for the problem.</u>
- Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.
- All chlorates are soluble, so AgClO₃ is soluble.
- Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.
Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.
Then, the precipiate will permit to conclude which flask contains AgCl.
Answer:
2Na + 2H2O - - - > 2NaOH + H2
Explanation: This is a an exothermic reaction.
Exothermic reaction : a reaction in which heat is released during a reaction with the product
Answer:
The average atomic mass of given element is 20.18 amu.
Explanation:
Given data:
Abundance of 1st isotope mass 20 amu = 90.92%
Abundance of 2nd isotope mass 21 amu = 0.257%
Abundance of 3rd isotope mass 22 amu = 8.82%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (90.92×20)+(0.257×21)+(8.82×22) /100
Average atomic mass = 1818.4 + 5.397 +194.04 / 100
Average atomic mass = 2017.819 / 100
Average atomic mass = 20.18 amu.
The average atomic mass of given element is 20.18 amu.