Answer:
Car B has more inertia than Car A
Explanation:
Given that,
Mass of car A = 1500 kg
Mass of car B = 2000 kg
Inertia is directly proportional to the mass of an object. Inertia is the measure of the mass of an object.
In this case, the mass of car B is more than that of car A, it means the inertia of car B is more than that of car A.
Hence, the correct option is (c) "Car B has more inertia than Car A".
Answer: 2.3 moles
Explanation:
Recall that based on Avogadro's law, 1 mole of any substance has 6.02 x 10^23 atoms
So if 1 mole of Aluminum = 6.02 x 10^23 atoms
Then, Z moles = 1.4 x 10^24 atoms
To get the value of Z, we cross multiply:
1 mole x 1.4 x 10^24 atoms = Z x (6.02 x 10^23 atoms)
1.4 x 10^24 atoms = Z x (6.02 x 10^23)
Hence, Z = (1.4 x 10^24 atoms) ➗ (6.02 x 10^23 atoms)
Z =2.3 moles
Thus, there are 2.3 moles in 1.4 x 10^24 atoms of aluminum.
Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
Because when you take the sand out it doesn't stay the same shape
Explanation:
The given data is as follows.
Now, according to Michaelis-Menten kinetics,
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
where, S = substrate concentration = 
M
Now, putting the given values into the above formula as follows.
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%20%5B%5Cfrac%7B10.4%20%5Ctimes%2010%5E%7B-6%7D%20M%7D%7B%2810.4%20%5Ctimes%2010%5E%7B-6%7DM%20%2B%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M%29%7D%5D)
= 
This means that 
would approache 
.
