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
Answer:
Oxygen.
Explanation:
The copper must be combined with something in the air.
Answer:
Atoms must have similar electronegativities in order to share electrons in a covalent bond.
Explanation:
Covalent bonding is one of the bondings that occurs between the atoms of elements. It is the bonding in which atoms share their valence electrons with one another. However, the ELECTRONEGATIVITY, which is the ability of an atom to be attracted to electrons play a major role in the formation of covalent bonds.
When atoms of different electronegativities combine, the more electronegative atom pulls more electrons towards itself, hence, an IONIC bond is formed. However, when the electronegativities of the atoms are similar, the sharing of their electrons becomes stronger. Hence, ATOMS MUST HAVE SIMILAR ELECTRONEGATIVITIES in order to share electrons in a covalent bond.
Answer:
73.88 g/mol
Explanation:
For this question we have to keep in mind that the unknown substance is a <u>gas</u>, therefore we can use the <u>ideal gas law</u>:

In this case we will have:
P= 1 atm
V= 3.16 L
T = 32 ªC = 305.15 ºK
R= 0.082 
n= ?
So, we can <u>solve for "n"</u> (moles):



Now, we have to remember that the <u>molar mass value has "g/mol"</u> units. We already have the grams (9.33 g), so we have to <u>divide</u> by the moles:


Answer:
RADIATION <em>is</em><em> </em><em>the</em><em> </em><em>transfer</em><em> </em><em>of</em><em> </em><em>Heat</em><em> </em><em>energy</em><em> </em><em>through</em><em> </em><em>space</em><em> </em><em>by</em><em> </em><em>Electromagnetic</em><em> </em><em>Radiation</em><em>.</em><em> </em><em>Most</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>electromagnetic</em><em> </em><em>radiation</em><em> </em><em>that</em><em> </em><em>comes</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>earth</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>Sun</em><em> </em><em>is</em><em> </em><em>invisible</em><em>.</em>