b. o2
oxygen is diatomic because 1 molecule is made of 2 atoms of oxygen
<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.
Answer:
The greenhouse effect is a phenomenon of radiative transfer, the process by which the energy of light waves is exchanged in matter. Radiative transfer dictates what energy is reflected, absorbed, and emitted. The greenhouse effect: A summary of the heat transfer in the Earth's atmosphere
Explanation:
trust me i have a huge brain and access to the internet
Answer:
Al(NO₃)₃ > KI > HF > CH₃OH
Explanation:
The electrical conductivities of the solutions will depend on the concentration of ions in solution.
Al(NO₃)₃ solution contains 0.1 M of Al³⁺ ions and 0.3 M of NO₃⁻ ions
KI solution contains 0.1 M of K⁺ ions and 0.1 M of I⁻ ions
HF solution contains less than 0.1 M of H⁺ ions and less then 0.1 M of F⁻ ions, because the HF acid will not dissociate completely
CH₃OH practically it does not dissociate, so in the solution will not be electrical conductive (comparative with the other solutions)
The solutions in order of decreasing intensity of the bulb are ranked as following:
Al(NO₃)₃ > KI > HF > CH₃OH
Withdrawal is the answer to your question.
