First, the sugar dissolves into small sugar molecules, Then fills in the spaces in between the water molecules thus, making Sugar water.
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The correct answer for the question that is being presented above is this one: "-2, -1, 0, 1, 2." T<span>he possible values of ml for an electron in a d orbital are -2, -1, 0, 1, 2. </span>Since the allowed values for mℓ range from −ℓ to +ℓ, once you know the value for ℓ you know the values for mℓ."
Answer:
2AgNO3 + Cu = Cu(NO3)2 + 2Ag
Explanation:
First you see which side has the most elements. Which is Cu(NO3)2 + Ag. But, both sides have the same elements? But, on the reactants side, there is 2 of NO3. On the products side there is only one.
Reactants: Products:
Cu = 1 Cu = 1
NO3 = 1 NO3 = 2
Ag = 1 Ag = 1
They are all equal, except for NO3. So on the reactants side, you add a two to make it even.
2AgNO3 + Cu = Cu(NO3)2 + Ag
Reactants: Products:
Cu = 1 Cu = 1
NO3 = 2 NO3 = 2
Ag = 2 Ag = 1
Now, the NO3 is equal, But the Ag isn't. But, you can add a 2 on the <u>products</u> side so the whole equation becomes equal.
2AgNO3 + Cu = Cu(NO3)2 + 2Ag
Reactants: Products:
Cu = 1 Cu = 1
NO3 = 2 NO3 = 2
Ag = 2 Ag = 2
They all have to do with science.
As given:
Initial moles of P taken = 2 mol
the products are R and Q
at equilibrium the moles of
R = x
total moles = 2 + x/2
Let us check for each reaction
A) P <-> 2Q+R
Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = 2x
moles of R = x
Total moles = (2-x) + 2x + x = 2 +2x
B) 2P <-> 2Q+R
Here x moles of P will give x moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x
moles of R = x/2
Total moles = (2-x) + x + x/2 = 2 + x/2
C) 2P <-> Q+R
Here x moles of P will give x/2 moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x /2
moles of R = x/2
Total moles = (2-x) + x + x = 2
D) 2P <-> Q+2R
Here x moles of P will give x/2 moles of Q and x moles of R
So at equilibrium
moles of P left = 2-x
moles of Q = x/2
moles of R = x
Total moles = (2-x) + x/2 + x = 2 + x/2
