If P=M*V than P=30kg*5m/s. P=150.
P=momentum
M=mass
V=Velocity
Now the last time i have done physics was last year. but i'm pretty confident in this answer. Hope this helps!
Both Physical and chemical changes.
Answer:
459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution
Explanation:
Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by the following expression:

Molarity is expressed in units 
In this case:
- Molarity: 1.56 M= 1.56

- Number of moles of calcium chlorine= ?
- Volume= 2.657 liters
Replacing:

Solving:
Number of moles of calcium chlorine= 1.56 M* 2.657 liters
Number of moles of calcium chlorine= 4.14 moles
In other side, you know:
- Ca: 40 g/mole
- Cl: 35.45 g/mole
Then the molar mass of the calcium chloride CaCl₂ is:
CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole
Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

mass= 459.126 g
<u><em>459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution</em></u>
Answer: But-2-enoic acid has
11 Sigma Bonds and
2 Pi Bonds.
Explanation: The sigma bonds which are formed due to head to head overlap of partally filled orbitals are shown in
red color, while Pi bonds which are formed after the formation of sigma bond by overlap of orbitals perpendicular to the sigma bond are shown in
blue color.
The average mass of an atom is calculated with the formula:
average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on
For the boron we have two isotopes, so the formula will become:
average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)
We plug in the values:
10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)
10.81 = 1.98 + 0.8020 × mass of isotope (2)
10.81 - 1.98 = 0.8020 × mass of isotope (2)
8.83 = 0.8020 × mass of isotope (2)
mass of isotope (2) = 8.83 / 0.8020
mass of isotope (2) = 11.009975
mass of isotope (1) = 10.012938 (given by the question)