The Andes was the mountain range that formed as a result of plate motion around the Ring of Fire.
Answer: The Andes
Atomic radius
Explanation:
A decrease in moving from the bottom of Group 3 to the top of the group shows an increase in trend from top to bottom.
The atomic radius of elements increases from top to bottom and decreases from the bottom to the top.
The elements in group 3 are B, Al , Ga, In and Tl
- Atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
- Downs a group from top to bottom, atomic radii increases progressively due to successive shells being added.
- This compensates for the size reducing effect of the increased nuclear charge.
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(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
Your answer would be 10 with the remainder of 1
(look at work for the work)