Answer:
4.65 L of NH₃ is required for the reaction
Explanation:
2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(s)
We determine the ammonium sulfate's moles that have been formed.
8.98 g . 1mol / 132.06 g = 0.068 moles
Now, we propose this rule of three:
1 mol of ammonium sulfate can be produced by 2 moles of ammonia
Therefore, 0.068 moles of salt were produced by (0.068 . 29) / 1 = 0.136 moles of NH₃. We apply the Ideal Gases Law, to determine the volume.
Firstly we do unit's conversions:
27.6°C +273 = 300.6 K
547.9 mmHg . 1 atm / 760 mmHg = 0.721 atm
V = ( n . R . T ) / P → (0.136 mol . 0.082 L.atm/mol.K . 300.6K) / 0.721 atm
V = 4.65 L
Answer:
Theoretical yield = 3.52 g
Percent yield =65.34%
Explanation:
Given data:
Mass of HgO = 46.8 g
Theoretical yield of O₂ = ?
Percent yield of O₂ = ?
Actual yield of O₂ = 2.30 g
Solution:
Chemical equation:
2HgO → 2Hg + O₂
Number of moles of HgO = mass/ molar mass
Number of moles of HgO = 46.8 g / 216.6 g/mol
Number of moles of HgO = 0.22 mol
Now we will compare the moles of HgO with oxygen.
HgO : O₂
2 : 1
0.22 : 1/2×0.22 = 0.11 mol
Theoretical yield:
Mass of oxygen = number of moles × molar mass
Mass of oxygen = 0.11 mol × 32 g/mol
Mass of oxygen = 3.52 g
Percent yield :
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.30 g/ 3.52 g × 100
Percent yield =65.34%
Answer: O D. A digital signal moves between a discrete number of values.
Explanation:
The digital signals are considered more reliable over the analog signals as they encode the information in a coded form. The bits or samples of the data are transmitted and converted into digital and numerical value. The stream of encoded data is in the form of continuous data at regular time intervals. It provides information in waveform and the data is in compact form. The data is in the form of binary bits 0 and 1 so greater the number of bits greater will be the greater will be the resolution of the information.
B. A gram would have a lot more molecules of propane than a mole
