weight depends on the force of gravity that is exerted on it.
<u>Answer:</u>
<em>20, 44, 62 </em>
<em></em>
<u>Explanation:</u>
To find the number of atoms of each element, we multiply coefficient and subscript
For example
contains
5 × 1 = 5 ,Ca atoms and
5 × 2 = 10, Cl atoms
If there is a bracket in the chemical formula
For example
we multiply coefficient × subscript × number outside the bracket to find the number of atoms
(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )
So
3 × 3 = 9 , Ca atoms
3 × 1 × 2 = 6, P atoms
3 × 4 × 2 = 24, O atoms are present.
So let us find the number of atoms of each element on the left side of the equation
![2C_{10} H_{22} (l)+31O_2 (g)\Rightarrow 20CO_2 (g)+22H_2 O(g)](https://tex.z-dn.net/?f=2C_%7B10%7D%20H_%7B22%7D%20%28l%29%2B31O_2%20%28g%29%5CRightarrow%2020CO_2%20%28g%29%2B22H_2%20O%28g%29)
Number of C atoms = 2 × 10 = 20
Number of H atoms = 2 × 22 = 44
Number of O atoms = 31 × 2 = 62
20, 44, 62 are the Answers.
Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
Answer:
Our world will never be the same. Our beaches will be submerged, many towns, cities, states, and even countries will be partially if not completely submerged. All these human cities being flooded will kill thousands of people, pollute the ocean, and lose billions of dollars.
Explanation:
Answer: d :The blue and orange soccer balls; they have more mass than the black soccer ball, but changed speed by the same amount.