Answer:
The pressure increases to 3.5 atm.
Solution:
According to Gay-Lussac's Law, " At constant volume and mass the pressure of gas is directly proportional to the applied temperature".
For initial and final states of a gas the equation is,
P₁ / T₁ = P₂ / T₂
Solving for P₂,
P₂ = P₁ T₂ / T₁ ----- (1)
Data Given;
P₁ = 3 atm
T₁ = 27 °C + 273 = 300 K
T₂ = 77 °C + 273 = 350 K
Putting values in eq. 1,
P₂ = (3 atm × 350 K) ÷ 300 K
P₂ = 3.5 atm
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
Answer:
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ is 4.11 × 10⁻³ grams
Explanation:
The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles
Therefore, we have;
One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂
3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂
The mass of one mole of Cl₂ = 70.906 g/mol
The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams
Therefore;
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ = 4.11 × 10⁻³ grams.
Metals, for example Cesium and Francium
