1-pentyne boiling point is 40 degrees C it is lower than the one for 1-octyne which is 126 degrees C the vapor pressure for 1-butene is higher at low temperatures than 1-heptene.These are due to the difference in the length of the chains. The strong molecular forces are stronger in large molecules. There is more energy needed to move large molecules to the vapor phase when in liquid form.
The two atoms shown in the equation are CALCIUM and oxygen.
<span>You start off with a neutral calcium atom with a shell of two electrons, a shell of 8 around that, a shell of 8 around that, and a shell containing 2...with no charge. </span>
<span>20 protons + 20 electrons. </span>
<span>You also have an oxygen atom with a shell of 2, and a shell of 6...with no charge. </span>
<span>8 protons + 8 electrons. </span>
<span>Each ionizes to form a calcium ion with 2 electrons removed (from the outer shell), leaving a +2 charge (20 protons, 18 electrons)... </span>
<span>and an oxygen ion with 2 electrons added (to the outer shell), leaving a -2 charge (8 protons, 10 electrons). </span>
<span>Their electrostatic attraction causes them to come together to form an ionic compound of CaO in a crystal lattice.</span>
<h3>
Answer:</h3>
0.387 J/g°C
<h3>
Explanation:</h3>
- To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
- Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
- That is, Q = mcΔT
in our question we are given;
Mass of copper, m as 95.4 g
Initial temperature = 25 °C
Final temperature = 48 °C
Thus, change in temperature, ΔT = 23°C
Quantity of heat absorbed, Q as 849 J
We are required to calculate the specific heat capacity of copper
Rearranging the formula we get
c = Q ÷ mΔT
Therefore,
Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)
= 0.3869 J/g°C
= 0.387 J/g°C
Therefore, the specific heat capacity of copper is 0.387 J/g°C
Answer:
An observation
Explanation: observation can also involve the perception and recording of data via the use of scientific instruments. The term may also refer to any data collected during the scientific activity
Answer:
The concentration of [Ca²⁺] is 8.47 x 10⁻³ M
Explanation:
We consider the solubility of hydroxyapatite,
Ca₁₀(PO₄)₆(OH)₂ ⇔ 10Ca²⁺ + 6PO₄³⁻ + 2 OH⁻
Assumed that there is <em>a</em> mol of hydoxyapatite disolved in water, yielding <em>10a</em> mol Ca²⁺ of and <em>6a</em> mol of PO₄³⁻
We also have Ksp equation,
Ksp = [Ca²⁺]¹⁰ x [PO₄³⁻]⁶ x [OH⁻]² = 2.34 x 10⁻⁵⁹
⇔ 10a¹⁰ x 6a⁶ x (5.30 x 10⁻⁶)² = 2.24 x 10⁻⁵⁹
⇔ 60a¹⁶ = 2.24 x 10⁻⁵⁹ / 5.30 x 10⁻¹²
⇔ a¹⁶ = 0.007 x 10⁻⁴⁷ = 7 x 10⁻⁵⁰
⇔ a = ![\sqrt[16]{7 . 10^{-50} }](https://tex.z-dn.net/?f=%5Csqrt%5B16%5D%7B7%20.%2010%5E%7B-50%7D%20%7D)
= 8.47 x 10⁻⁴
Hence,
[Ca²⁺] = 10<em>a</em> = 8.47 x 10⁻³ M
