<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.
Explanation:
Given parameters:
Wavelength of photon = 827nm = 827 x 10⁻⁹m
Unknown:
Energy of the photon = ?
Type of radiation = ?
Solution:
The energy of a photon can be derived using the expression below:
E = 
h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s
c is the speed of light = 3 x 10⁸m/s
Insert the parameters and solve;
E = 
E = 2.4 x 10⁻¹⁹J
Type of radiation:
Near infrared radiation
Water molecules are constantly moving in Ice they move slower and are more tightly packed together and move at a slower pace to lower the freezing point you would have to find a solute that makes it harder for water to form crystals
Answer:
0.260 Celsius
Explanation:
q =c x m x (T2-T1)
c - specific heat of water 4.186 J/g.C
T2-T1 = q /(c x m) = 0.959 /(4.186 x 0.88) = 0.959/3.68 =0.260 C
