Answer:Well There is many ways to increase friction between objects. One being rubbing the two objects together quicker and harder. If there is any sort of wetness or anything related to that make sure to dry the surface between the two objects you want to create friction between so it will be more effective.
Explanation:
Answer:
Physical change: A,C ,D, F, H, I
Chemical Change: B, E, G, J, K
Lithium i believe is the answer
<h3>
Answer:</h3>
2.809 L of H₂SO₄
<h3>
Explanation:</h3>
Concept tested: Moles and Molarity
In this case we are give;
Mass of solid sodium hydroxide as 13.20 g
Molarity of H₂SO₄ as 0.235 M
We are required to determine the volume of H₂SO₄ required
<h3>First: We need to write the balanced equation for the reaction.</h3>
- The reaction between NaOH and H₂SO₄ is a neutralization reaction.
- The balanced equation for the reaction is;
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
<h3>Second: We calculate the umber of moles of NaOH used </h3>
- Number of moles = Mass ÷ Molar mass
- Molar mass of NaOH is 40.0 g/mol
Moles of NaOH = 13.20 g ÷ 40.0 g/mol
= 0.33 moles
<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
- From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
- Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
- Thus, Moles of H₂SO₄ = moles of NaOH × 2
= 0.33 moles × 2
= 0.66 moles of H₂SO₄
<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
- When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
- That is; Volume = Number of moles ÷ Molarity
In this case;
Volume of the acid = 0.66 moles ÷ 0.235 M
= 2.809 L
Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.
In a redox chemical reaction, one species gets reduced and another gets oxidized. Manganese element is reduced in this reaction.
<h3>What is oxidized and reduced?</h3>
In a redox reaction, the increase or decrease in the oxidation number and electrons results in the reduction and oxidation of the chemical species. The oxidation and reduction occur simultaneously in a reaction.
The oxidation number of Mn in permanganate ion was +8 on the left side and decreased to +4 on the right side of the equation. Potassium permanganate is an oxidizing agent that has reduced the manganese ion of the permanganate ion.
Therefore, manganese is reduced.
Learn more about reduction and oxidation here:
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