1) Zn + 2 HCl = ZnCl2 + H2 ( <span>single replacement )
2) </span>2 NaCl + F2 = 2 NaF + Cl2 ( <span>single replacement )
3) </span>2 AlBr3 + 3 K2SO4 = 6 KBr + Al2(SO4)3 ( <span>double replacement )
4) </span>2 K + MgBr2 = 2 KBr + Mg ( <span>single replacement )
Answer 3
hope this helps!</span>
Answer:
N=2
H=6
Explanation:
1.Balance a chemical equation in terms of moles.
2.Use the balanced equation to construct conversion factors in terms of moles.
3.Calculate moles of one substance from moles of another substance using a balanced chemical equation.
The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products.
(P.s it could also be where you have to solve it in which you have to simplify it first then solve it.) like adding them all up.
Hope this is the answer. :)
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
