Answer:
Kinetic energy is directly proportional to mass
Explanation:
Kinetic energy is directly proportional to the mass of an object and also directly proportional to the square of the velocity of that object:
Notice that if we keep velocity constant and only increase the mass of a object, the kinetic energy of that object would increase, as we've already emphasized the direct relationship between the kinetic energy term and the mass term.
Let's take a simple example: assume that object 1 and object 2 are both moving at the same velocity but object 1 has a much lower mass than object 2. According to the equation, object 1 has lower kinetic energy. This object can then transform all of its kinetic energy into some other form, say, heat the ground. The heat transferred will be significantly lower than by the object 2 moving at the same velocity but having a much greater mass.
Answer:
The answer to your question is 0.269 g of Pb
Explanation:
Data
Lead solution = 0.000013 M
Volume = 100 L
mass = 0.269 g
atomic mass Pb = 207.2 g
Chemical reaction
2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)
Process
1.- Calculate the mass of Pb in solution
Formula
Molarity = 
Solve for number of moles
Number of moles = Volume x Molarity
Substitution
Number of moles = 100 x 0.000013
Number of moles = 0.0013
2.- Calculate the mass of Pb formed.
207.2 g of Pb ----------------- 1 mol
x g ----------------- 0.0013 moles
x = (0.0013 x 207.2) / 1
x = 0.269 g of Pb
Answer:
72.6 grams
Explanation:
I got this answer through stoichiometry. For every 1 mole of Mg, 2 moles of CuBr are consumed. Because of this, multiply the moles of Mg by 2. Then, convert moles to grams.
The scientist should make a new hypothesis and do a new experiment! That is what I say!
Answer:
Explanation:
Hello,
In this case, for the given reaction:
In such a way, the initial concentrations are:
![[H_2]_0=\frac{1.30x10^{-2}mol}{0.1L}=0.130M](https://tex.z-dn.net/?f=%5BH_2%5D_0%3D%5Cfrac%7B1.30x10%5E%7B-2%7Dmol%7D%7B0.1L%7D%3D0.130M)
![[NO]_0=\frac{2.60x10^{-2}mol}{0.1L}=0.260M](https://tex.z-dn.net/?f=%5BNO%5D_0%3D%5Cfrac%7B2.60x10%5E%7B-2%7Dmol%7D%7B0.1L%7D%3D0.260M)
Thus, at equilibrium, the change 
, due to the chemical reaction extent, turns out:
![x=\frac{[NO]_{0}-[NO]_{eq}}{2} =\frac{0.260M-0.161M}{2}=0.049M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BNO%5D_%7B0%7D-%5BNO%5D_%7Beq%7D%7D%7B2%7D%20%3D%5Cfrac%7B0.260M-0.161M%7D%7B2%7D%3D0.049M)
Thus, the rest of the concentrations at equilibrium are:
![[H_2]_{eq}=0.130M-2(0.049M)=0.032M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.130M-2%280.049M%29%3D0.032M)
![[H_2O]_{eq}=2(0.049M)=0.098M](https://tex.z-dn.net/?f=%5BH_2O%5D_%7Beq%7D%3D2%280.049M%29%3D0.098M)
![[N_2]_{eq}=0.049M](https://tex.z-dn.net/?f=%5BN_2%5D_%7Beq%7D%3D0.049M)
In such a way, the equilibrium constant for the reaction, result as follows, even when on the statement the NO is excluded, because it participates in the equilibrium:
![Kc=\frac{[H_2O]_{eq}^2[N_2]_{eq}}{[H_2]_{eq}^2[NO]_{eq}^2}=\frac{(0.098M)^2(0.049M)}{(0.032M)^2(0.161M)^2} \\\\Kc=17.7](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D_%7Beq%7D%5E2%5BN_2%5D_%7Beq%7D%7D%7B%5BH_2%5D_%7Beq%7D%5E2%5BNO%5D_%7Beq%7D%5E2%7D%3D%5Cfrac%7B%280.098M%29%5E2%280.049M%29%7D%7B%280.032M%29%5E2%280.161M%29%5E2%7D%20%5C%5C%5C%5CKc%3D17.7)
Best regards.
