Answer:
V₂ = 0.95 L
Explanation:
Given data:
Initial temperature of gas = 171.4 K
Final temperature of gas = 288.4 K
Final volume = 1.6 L
Initial volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 1.6 L × 171.4 K / 288.4 k
V₂ = 274.24 L.K / 288.4 K
V₂ = 0.95 L
Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
The primary structure is the amino acids' unique sequence. The polypeptide's local folding to form structures such as the α-helix and β-pleated sheet constitutes the secondary structure. The overall three-dimensional structure is the tertiary structure
Answer:
light bends and makes effects in the water
Explanation:
Sodium reacts to chlorine and gives NaCl. The balanced reaction is given below:
2Na + Cl₂→ 2NaCl. Two moles Na reacts with one mole Cl₂ and produces two moles of NaCl. Atomic mass of Na= 23, Molar mass of Cl₂= 71, molar mass of NaCl=58.5.
So, 46 g Na reacts with 71 g of Cl₂ and produces (2 X 58.5)g = 117 g of NaCl. As per question Na reacts completely which means Na is the limiting reagent. So, number of moles of Na reacts = number moles of NaCl produced.
NaCl produced= 819 g= (819/58.5) moles= 15.69 moles. Therefore, 15.69 moles = 15.69 X 23 g=360.87 g of Na reacted.
