It is estimated that about 1 ft of snow is equal to 1 inch of rain.
Converting inch to cm
1 inch=2.54 cm
Thus, 1 cm=0.393 inch
2.5 cm of rain will be 0.984 inch
now 1 inch of rain is equal to 1 ft of snow, thus, 0.983 inch of rain will be:
0.983 inch of rain=0.983 ft of snow.
Converting ft to cm:
1 ft =30.48 cm
Thus,
0.983 ft= 29.96 cm
Thus, 2.5 cm of rain= 29.96 cm of snow.
Answer:
1.68 × 10²³ Molecules
Explanation:
As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (molecules) contained by 0.280 moles of Br₂, we will use following relation,
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of Molecules,
Number of Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting values,
Number of Molecules = 0.280 mol × 6.022 × 10²³ Molecules.mol⁻¹
Number of Molecules = 1.68 × 10²³ Molecules
Hence,
There are 1.68 × 10²³ Molecules present in 0.280 moles of Br₂.
Answer:
- <u><em>34 g of NH₃ </em></u><em>will be produced from the reaction of 28 g of N₂ with 25 g of H₂.</em>
Explanation:
1) <u>Balanced chemical equation</u>
- N₂ (g) + 3H₂ (g) → 2NH₃(g)
2) <u>Stoichiometric (theoretical ) mole ratios</u>
- 1 mol N₂ (g) : 3mol H₂ (g) : 2 mol NH₃(g)
3) <u>Number of moles of each reactant</u>
- number of moles = mass in grams / atomic mass
- number of moles of N₂ = 28 g / 28 g/mol = 1 mol
- number of moles of H₂: 25 g / 2 g/mol = 12.5 mol
4)<u> Limiting reactant</u>
Since the stoichiometry states that 1 mol of N₂ reacts with 3 moles of H₂, the given mass of N₂ will react completely with the given amount of H₂, and the calculations must be done with the 28 g (1 mol) of N₂ as the limiting reactant.
5) <u>Yield</u>
Set the proportion with the mole ratios:
1 mol H₂ / 2 mol NH₃ = 1 mol H₂ / x ⇒ x = 2 mol NH₃
6) <u>Convert to grams</u>
- mass in grams = number of moles × molar mass = 2 mol × 17 g/mol = 34 g.
Answer: <em>the reaction of 28 g of N₂ with 25 g of H₂ will produce 34 g of NH₃</em>
Then the liqid would freeze. Freezeing point is the tempture of a liqid that would make it liquid.