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The answer is in the photo.
<span>(2.09 mL) x (1.592 g/mL) / (227.0871 g C3H5O9N3/mol) = 0.014652 mole C3H5O9N
4 moles C3H5O9N produce 12 + 6 + 1 + 10 = 29 moles of gases, so:
(0.014652 mole C3H5O9N) x (29/4) = 0.106 mole of gases
(b)
(0.106 mol) x (46 L/mol) = 4.88 L gases
(c)
(0.014652 mole C3H5O9N) x (6/4) x (28.0134 g/mol) = 0.616 g N2</span>
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
The answer is: True.
The magnitude of a vector is represented by the length of the arrow.
The arrow length is drawn according a chosen scale.
For example, the diagram shows a vector with a magnitude of 100 kilometers, if the scale used for constructing the diagram is 1 cm = 10 km, the vector arrow is drawn with a length of 10 cm.
The arrow has an obvious tail and arrowhead. The arrow points in the precise direction.
