Answer:
Answers with detail are given below
Explanation:
1) Given data:
Mass of Rb₃Rn = 76.19 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 478.43 g/mol
Number of moles = 76.19 g/ 478.43 g/mol
Number of moles = 0.16 mol
2) Given data:
Mass of FrBi₂ = 120.02 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 640.96 g/mol
Number of moles = 120.02 g/640.96 g/mol
Number of moles = 0.19 mol
3) Given data:
Mass of Zn₂F₃ = 88.24 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 187.73 g/mol
Number of moles = 88.24 g/ 187.73 g/mol
Number of moles = 0.47 mol
4) Given data:
Number of moles of Sb₄Cl = 1.20 mol
Mass of Sb₄Cl = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 522.49 g/mol
Mass = Number of moles × molar mass
Mass = 1.20 mol × 522.49 g/mol
Mass = 626.99 g
Answer:
Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.
The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.
Answer:
What datatable? Picture please!
Explanation:
Answer:
55.18 L
Explanation:
First we convert 113.4 g of NO₂ into moles, using its molar mass:
- 113.4 g ÷ 46 g/mol = 2.465 mol
Then we<u> use the PV=nRT formula</u>, where:
- P = 1atm & T = 273K (This means STP)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
Input the data:
- 1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K
And <u>solve for V</u>:
<span>Catalysts decrease the activation energy and the more collisions result in a </span>reaction<span>, so the </span>rate<span> of </span><span>reaction increases.</span><span />
