Answer:
the answer is D
Explanation:
the reason how ik this is that i already did this test so yea
Answer: 209.2
Explanation:
4. q = mcat
q = 5g(4.184)(40-50)
q = -209.2
the negative sign indicates that heat is given off
Answer:
see explanation below
Explanation:
In this case, we have the following reaction:
Mg + 1/2O₂ -------> MgO
Now, according to this reaction we want to know the percent composition of MgO. The problem is not providing the mass of the initial reactants and the product, so we can use the atomic weights of the components, to do this
The molecular weight of Mg is 24.305 g/mol, and O is 15.999 g/mol, so, let's calculate the molar mass of MgO:
MM MgO = 24.305 + 15.999 = 40.304 g/mol
Now with this weight, let's see the percent composition of this compound:
%Mg = 24.305 / 40.304 * 100 = 60.304 %
%O = 15.999 / 40.304 * 100 = 39.696 %
And this would be the percent composition of MgO
Answer: 3.0 kJ × 1 mol/40.65 kJ× 18.02 g/mol × 1 mL/1 g= 1.3 mL
Answer:
Explanation:
Hello,
The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:
![\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Now, we answer:
(a)
![\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D%3D4%2A%5Cfrac%7B1%7D%7B-3%7D%20%2A%28-0.12M%2Fs%29%3D0.16M%2Fs)
(b)
![\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%3D-1%2A%5Cfrac%7B1%7D%7B5%7D%20%2A%280.2M%2Fs%29%3D-0.04M%2Fs)
(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:
![r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B-3%7D%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-1%7D%20%5Cfrac%7Bd%5BE%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B-2%7D%20%5Cfrac%7Bd%5BF%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cfrac%7Bd%5BG%5D%7D%7Bdt%7D%20%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7Bd%5BH%5D%7D%7Bdt%7D)
Thus, any of the available expressions are suitable to quantify the rate of the reaction.
Best regards.