Answer:
The question is incomplete. The response options are as follows:
I. C-O or C=O
II. C-C
III. C-H
IV. O-H
The answer is: IV>III>I>II
Explanation:
V) O-H is a hydrogen bridge. The hydrogen bridge is characterized by being similar to dipole-dipole bonds.
III) C-H is an ionic bond. The ionic bond occurs when they fuse together due to electron transfer.
I) C=O is a covalent bond. The covalent bond happens when two atoms bond together to create a molecule, sharing its electrons that are in its most superficial layer,
II) C-C is covalent bond.
4NH3 + 5O2 ==> 4NO + 6H2O Balanced equation
ALWAYS WORK IN MOLES, NOT IN GRAMS
moles of NO produced = 70.5 g NO x 1 mole/30 g = 2.35 moles NO
Since this represents only a 29.8% yield, find what 100% yield would be:
2.35 moles/0.298 = 7.89 moles of NO
From the balanced equation 4 moles NH3 produces 4 moles of NO. Calculate moles of NH3 needed:
7.89 moles NO x 4 moles NH3/4 moles NO = 7.89 moles NH3 needed
Find grams of NH3 needed:
7.89 moles NH3 x 17 g/mole = 134 g NH3 needed
Answer:
2192.64 PSI.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the container in L (V = 1650 L).
n is the no. of moles of the gas in mol (n = 9750 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature of the gas in (T = 35°C + 273 = 308 K).
∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.
- <u><em>To convert from atm to PSI:</em></u>
1 atm = 14.696 PSI.
<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3