Answer:
4.083 * 10^20 atoms.
Explanation:
One Mole of phosphorus contains 6.022 * 10^23 atoms (Avogadros number)'
Since 1 mole of Phosphorus has a mass of 30.974 grams, 21 milligrams has
6.022 * 10^23 * 0.021 / 30.974
= 0.004083 * 10^23
= 4.083 * 10^20
Sugar, sucrose (C12H22O11: a disaccharide, composed of the two monosaccharides: glucose and fructose), is odorless, that is, it lacks odor. When heated a phase change occurs resulting in melting of a thick syrup.
The empirical formula is a formula of a compound showing the proportion of each element involved in the compounds but it does not represent the total number of atoms in the compound. It is the lowest number of ratio between the elements in the compound. In order, to determine the actual number of the atoms or the molecular formula of the compounds, we make use of the molar mass of the compound.
<span>To
determine the molecular formula, we multiply a value to the empirical formula.
Then, calculate the molar mass and see whether it is equal to the one
given (104.1 g/ mol). From the choices, the only valid options are b, d and e.
</span> molar mass
1 CH 13.02
8 C8H8 104.16
6 C6H6 78.12
Therefore the correct answer is option B.
Answer:
43.868 J
Explanation:
Kinetic energy of a body is the amount of energy possessed by a moving body. The SI unit of kinetic energy is the joule (kg⋅m²⋅s⁻²).
According to classical mechanics, kinetic energy = 1/2 m·v²
Where, m= mass in kg and v= velocity in m/s
Given: m = 19.2 lb and v = 7.10 miles/h
Since, 1 lb= 0.453592 kg
∴ m = 19.2 lb = 19.2 × 0.453592 kg = 8.709 kg
Also, 1 mi = 1609.34 m and 1 h = 3600 sec
∴ v = 7.10 mi/h = 7.10 × 1609.34 m ÷ 3600 sec = 3.174 m/sec
Therefore, <u>kinetic energy of the goose</u> = 1/2 m·v² = 1/2 × (8.709 kg)× (3.174 m/sec)² = 43.868 J
Answer:
The law is observed in the given equation.
Explanation:
CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂
In order to find out if the law of conservative mass is followed, we need to <u>count how many atoms of each element are there in both sides of the equation</u>:
- Ca ⇒ 1 on the left, 1 on the right.
- C ⇒ 1 on the left, 1 on the right.
- O ⇒ 3 on the left, 3 on the right.
- H ⇒ 2 on the left, 2 on the right.
- Cl ⇒ 2 on the left, 2 on the right.
As the numbers for all elements involved are the same, the law is observed in the given equation.
