<span>C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.
heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°
final temp = 21 + 14° = 35°C</span>
Answer:
270. mL
General Formulas and Concepts:
<u>Acid-Base Titrations</u>
Dilution: M₁V₁ = M₂V₂
- M₁ is stock molarity
- V₁ is stock volume
- M₂ is new molarity
- V₂ is new volume
Explanation:
<u>Step 1: Define</u>
<em>Identify</em>
[Given] V₂ = 629.1 mL
[Given] M₂ = 11.2 M
[Given] M₁ = 26.1 M
[Solve] V₁
<u>Step 2: Find Stock Volume</u>
- Substitute in variables [Dilution]: (26.1 M)V₁ = (11.2 M)(629.1 mL)
- Multiply: (26.1 M)V₁ = 7045.92 M · mL
- Isolate V₁ [Cancel out units]: V₁ = 269.959 mL
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs as our lowest.</em>
269.959 mL ≈ 270. mL
<span>The latent heat of vaporization refers to the amount of heat that must be added to a liquid substance in order to transform it into gas. </span>It has the effect of cooling a liquid because in the process of vaporizing the liquid, heat is absorbed by the molecules which is used to change their phase from liquid to gas. As a result, since heat has been taken away by the process of vaporization, the remaining liquid becomes cooler.
Answer:
4.81×10¹⁰ atoms.
Explanation:
We'll begin by converting 3.2 pg to Ca to grams (g). This can be obtained as follow:
1 pg = 1×10¯¹² g
Therefore,
3.2 pg = 3.2 pg × 1×10¯¹² g / 1 pg
3.2 pg = 3.2×10¯¹² g
Therefore, 3.2 pg is equivalent to 3.2×10¯¹² g
Next, we shall determine the number of mole in 3.2×10¯¹² g of Ca. This can be obtained as follow:
Mass of Ca = 3.2×10¯¹² g
Molar mass of Ca = 40.08 g/mol
Mole of ca=.?
Mole = mass /molar mass
Mole of Ca = 3.2×10¯¹² / 40.08
Mole of Ca = 7.98×10¯¹⁴ mole.
Finally, we shall determine the number of atoms present in 7.98×10¯¹⁴ mole of Ca. This can be obtained as illustrated below:
From Avogadro's hypothesis,
1 mole of Ca contains 6.02×10²³ atoms.
Therefore, 7.98×10¯¹⁴ mole of Ca will contain = 7.98×10¯¹⁴ × 6.02×10²³ = 4.81×10¹⁰ atoms.
Therefore, 3.2 pg of Ca contains 4.81×10¹⁰ atoms.