Explanation: Hydrogen bonds are the strongest one of the intermolecular forces. A hydrogen bond is a bond between hydrogen in one molecule or the other ones are fluorine and nitrogen. So it's between a hydrogen in one molecule and an electronegative atom in another molecule. So they always involve hydrogen.
Answer:
The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.
The time needed for x oil changes is 12 * x.
The time needed for y tire changes is 18 * y.
The total time is the sum of the above times and needs to be less than 180 that is
12 * x + 18 * y < 180 divide both sides of equation by 6
12/6 * x + 18/6*y < 180/6
2*x + 3*y < 30
2*x < 30 - 3*y divide both sides by 2 to get the inequality for x
x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15
2*x + 3*y < 30
3*y < 30 - 2*x divide both sides by 3 to get the inequality for y
y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10
Also we can write x + y < x+ 3/2 * y < 15.
Explanation:
Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Answer:
The total pressure is 27.8 atm
Explanation:
From the ideal gas equation,
PV = nRT
P (total pressure) = nRT/V
n (total moles of gases) = (6/1 moles of hydrogen) + (15.2/14 moles of nitrogen) + (16.8/4 moles of helium) = 6+1.1+4.2 = 11.3 moles
R = 0.082057L.atm/gmol.K, T = 27°C = 27+273K = 300K, V = 10L
P = 11.3×0.082057×300/10 = 27.8 atm
