Answer:
The molar mass of the gas is 36.25 g/mol.
Explanation:
- To solve this problem, we can use the mathematical relation:
ν = 
Where, ν is the speed of light in a gas <em>(ν = 449 m/s)</em>,
R is the universal gas constant <em>(R = 8.314 J/mol.K)</em>,
T is the temperature of the gas in Kelvin <em>(T = 20 °C + 273 = 293 K)</em>,
M is the molar mass of the gas in <em>(Kg/mol)</em>.
ν =
(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,
<em>by squaring the two sides:</em>
(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,
∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.
<em>∴ The molar mass of the gas is 36.25 g/mol.</em>
Answer:
50 N ( left)
Explanation:
Given data:
The force on right side = 450 N
The force on left side = 500 N
Net force = ?
Solution:
F (net) = 500 N - 450 N
F (net) = 50 N ( left)
V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles = 
Number of moles = 
= 0.003mole
*A & B*
Answers A & B are not possible, as Hydrogen “bonds” are intermolecular forces and do not actually involve transfer or sharing of electrons.
*C & D*
Viscosity and surface tension are not the answer as they are not specific enough to the question.
*E*
Polarity of water molecules is the correct answer, as water molecules are highly polar. The partial positive of the Hydrogen on one water molecule is highly attracted to the partial negative of the Oxygen (due to its lone pairs) on another water molecule.
