Answer:
a. NH3 is limiting reactant.
b. 44g of NO
c. 40g of H2O
Explanation:
Based on the reaction:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.
To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:
<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>
25g NH3*(1mol/17.031g) = 1.47moles NH3
Moles O2 = 4 moles
For a complete reaction of 4 moles of O2 are required:
4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.
As there are just 1.47 moles, NH3 is limiting reactant
b. Moles NO:
1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO
Mass NO -Molar mass: 30.01g/mol-
1.47mol NO * (30.01g/mol) = 44g of NO
c. Moles H2O:
1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O
Mass H2O -Molar mass: 18.01g/mol-
2.205mol H2O * (18.01g/mol) = 40g of H2O
A: making s sandcastle. This is because water and sand is only a mixture, so they do not react with each other. All the rest include chemical reactions!
Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
The large piece of jewelry that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml
To solve this problem the formulas and the procedures that we have to use are:
Where:
- d= density
- m= mass
- v= volume
- v(f) = final volume
- v(i) = initial volume
Information about the problem:
- m = 132.6 g
- v(i) = 48.6 ml
- v(f) = 61.2 ml
- v = ?
- d =?
Applying the volume formula we get:
v = v(f)-v(i)
v = 61.2 ml - 48.6 ml
v = 12.6 ml
Applying the density formula we get:
d = m/v
d = 132.6 g/12.6 ml
d = 10.523 g/ml
<h3>What is density?</h3>
It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.
Learn more about density in: brainly.com/question/1354972
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