I believe the answer is 80. I am not sure
Answer:
The speed of the particle at x = 4.0 m is 13.66 m/s
Explanation:
The work done by this force between the two points above is given by
W = ∫ F dx
W = ∫⁴₋₄ (-5x² + 7x) dx
W = [(-5x³/3) + (7x²/2)]⁴₋₄
W = [(-5(4³)/3) + (7(4²)/2)] - [(-5(-4)³)/3) + (7((-4)²)/2)] = (-50.6667) - (162.6667) = (- 213.33 J)
Kinetic energy at -4.0 m
At this point, v = 20 m/s
K.E = mv²/2 = 2 × 20²/2 = 400 J
To obtain the kinetic energy at 4 m,
We apply the work-energy theorem which mathematically translates to
The work done in moving a particle from one point to another = Change in kinetic energy of the particle between those two points
W = ΔK.E
Work done between x = - 4m and x = 4 m is - 213.33 J
Hence, ΔK.E = -213.33 J
Change in kinetic energy of the particle between x = - 4m and x = 4m is ΔK.E
ΔK.E = (Kinetic energy at x = 4m) - (kinetic energy at x = - 4m)
- 213.33 J = (mv²/2) - 400
mv²/2 = -213.33 + 400 = 186.67 J
2v² = 2 × 186.67
v² = 186.67
v = 13.66 m/s.
<span>54.8 g of MgI2 can be produced.
To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium
Atomic weight of Iodine = 126.90447
Atomic weight of Magnesium = 24.305
Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394
Now determine how many moles of Iodine and Magnesium you have
moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole
moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole
Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent.
So figure out how many moles of magnesium will be consumed by the iodine
0.393997154 mole / 2 = 0.196998577 mole.
This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2
0.196998577 mole * 278.11394 g/mole = 54.78805 g
Round the result to the correct number of significant figures.
54.78805 g = 54.8 g</span>
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
Answer:
Given : No. Of moles = 1.5
To calculate : no. Of molecules =N
We know that moles = N / 6.022 x 10²³
Therefore, 1.5 x 6.022 x 10²³ = N
Hence N = 9.0330x 10²³ molecules
