D is to stop the current and the force can be removed
Okay Elements can combine to created element compounds a few examples can be lets say Gold(Au) Is an example of an Iconic Compound (usually a common compound). An Element Mixture is literal a mixture Ex. Brass. Brass is usually 10% copper and 45% zinc. So in order to get Brass you must physically m<span />ix copper with zinc and the other metal (sorry i don't remember the name) Also mixtures tend to not be chemically constant, this is because it was physically mixed rather than all of the atoms be the same if they were chemically combined.
You want to convert Euros / kg to Dollars / pound
So you'll have to multiply the price (Euros / kg) by the exchange rate (Dollars / Euro) to get a number that has units of Dollars / kg.
Euros Dollars Dollars
------- x --------- = --------
kg Euro kg
Then you have to multiply the number of kg in a pound (which has units of kg / pound) to get a number in units of Dollars / pound. This number is 2.2.
Dollars kg Dollars
-------- x -------- = --------
kg pound pound
So the answer would be 1.75 * 1.36 / 2.2 = 1.08 dollars / pound.
Answer:
0.364
Explanation:
Let's do an equilibrium chart for the reaction of combustion of ammonia:
2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)
4.8atm 1.9atm 0 0 Initial
-2x -(3/2)x +x +3x Reacts (stoichiometry is 2:3/2:1:3)
4.8-2x 1.9-(3/2)x x 3x Equilibrium
At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.
If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium
pNH₃ = 4.8 - 2*0.63 = 3.54 atm
pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm
pH₂O = 3*0.63 = 1.89 atm
The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:
Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*
]
Kp = [0.63*(1.89)³]/[(3.54)²*
]
Kp = 4.2533/11.6953
Kp = 0.364
Answer:
1.- Chemical change
2.- Because the atoms in the image are tranformed into a new molecule
3.- Yes, it does
4.- Due to the amount of atoms are the same both in products and reagents
Explanation:
