Atom can never be divided into smaller particles is the answer was disproved
Answer:
a. 9.2
b. 4.4
c. 6.3
Explanation:
In order to calculate the pH of each solution, we will use the definition of pH.
pH = -log [H⁺]
(a) [H⁺] = 5.4 × 10⁻¹⁰ M
pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2
Since pH > 7, the solution is basic.
(b) [H⁺] = 4.3 × 10⁻⁵ M
pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4
Since pH < 7, the solution is acid.
(c) [H⁺] = 5.4 × 10⁻⁷ M
pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3
Since pH < 7, the solution is acid.
Answer: A) Inconclusive; you would not know which of the two variables caused the change.
Explanation:
When you set up an experiment, you must make sure that you control the variables such that only one independent variable changes at a time, while all the remainder conditions (the other independent variables) are controlled (fixed).
By observing (measuring) the dependent variable, while only one independent variable changes you can understandhow such independent variable explains (determines) the dependent variable, leading to a conclusion.
Conversely, if two or more independent variables change at a time, then there is no way that you can tell how the output (dependent variable) is related with one or other of the changes of the indipendent variables. You wolud not be able to discriminate (distinguish) the effect of one or other variable, making the experiment inconclusive
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Answer:
Explanation:
1. Write the skeleton equation for the half-reaction
NO₃⁻ ⟶ N₂O
2. Balance all atoms other than H and O
2NO₃⁻ ⟶ N₂O
3. Balance O by adding H₂O molecules to the deficient side.
2NO₃⁻ ⟶ N₂O + 5H₂O
4. Balance H by adding H⁺ ions to the deficient side.
2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O
5. Balance charge by adding electrons to the deficient side.
2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O
The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F
<span>2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045
moles x = 1.68/ 55.9=0.03
0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O
to get whole numbers multiply by 2
x2O3
X2O3 +3 CO = 2 X + 3 CO2</span>
