Freezing, condensation, Deposition.
Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:
First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?
= 25000 g
Substitute the values into the formula:
= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
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.
Answer:
A
Explanation:
heat energy is transfered through a hot object touching a cold object
__ KClO₃ → __ KCl + __ O₂
Left Side:
1 K
1 Cl
3 O
Right Side:
1 K
1 Cl
2 O
Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.
This gives us 2KClO₃ → __ KCl + 3O₂.
However, this equation is still not balanced.
Left Side:
2 K
2 Cl
6 O
Right Side:
1 K
1 Cl
6 O
In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.
2KClO₃ → 2KCl + 3O₂
