Arietis, camelopardalis, lyncis, scorpii, cancri, canis minoris
Answer:
9.82 g of Mg(NO₃)₂
Explanation:
Let's determine the reaction:
2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr
2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.
We determine the moles of AgNO₃
22.5 g . 1mol / 169.87g = 0.132 moles
Ratio is 2:1.
2 moles of silver nitrate can produce 1 mol of magnesium nitrate
Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles
We convert moles to mass:
0.0662 mol . 148.3 g/ mol = 9.82 g
Answer: Heat causes surrounding air to rapidly expand and vibrate, which creates the pealing thunder we hear a short time after seeing a lightning flash.
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
Answer:
+1.03 V
Explanation:
The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).
The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:
Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V
Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V
The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.
emf = E°reduces - E°oxides
emf = 0.80 - (-0.23)
emf = +1.03 V
