The percentage adds up to 100%.
<h3>Percentages</h3>
Mass of mixture = mass of container+mixture - mass of container = 56.779 - 54.558 = 2.221 g
CuCO3 is insoluble in water. Thus:
Mass of CuCO3 = mass of beaker and residue - mass of beaker = 78.875 - 77.575 = 1.300 g
Mass of NaCl = mass of mixture - mass of CuCO3 = 2.221 - 1.300 = 0.921 g
%NaCl (w/w) = weight of NaCl/weight of mixture = 10.921/2.221 = 41.468%
% (w/w) CuCO3 = weight of CuCO3/weight of mixture = 1.300/2.221 = 58.532%
Addition of percentages = 41.468 + 58.532 = 100%
More on percentages can be found here: brainly.com/question/24159063
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2H2 + O2 ---->2H2O
number of moles in reaction 2 mol 1 mol 2 mol
number of liters in the reaction 2*22.4 L 1*22.4 L 2*22.4L
We can see that volumes of the gases are proportional to coefficients in the reaction ( if gases are under the same conditions), so we can write
2H2 + O2 ---->2H2O
2 L 1 L 2 L
given 40 L ( 25 L) 40 L
We can see that we have excess of O2,
because if 2 L H2 are needed 1 L O2, then 40 L of H2 are needed 20 L O2.
So, limiting reactant is H2, and we will need to calculate Volume of H2O using H2.
2L H2 give 2L H2O(gas), so 40 L H2 give 40 L H2O.
Answer:
10 atm
Explanation:
There's a lot to do here, but lets take it one step at a time. First, let's write a balanced equation for the decomposition of potassium chlorate into potassium chloride and oxgyen gas.
2 KClO3 → 2 KCl + 3 O2
Now let's find the moles of the KClO3 (molar mass 122.55 g/mol) that we have take 10 g/122.55 g/mol, grams will cancel and we are left with 0.0816 moles. lets divide that by two since we have a two in front of the KClO3 in the equation, and then multiply that number by 5 since it's the total moles of products, in summary, multiply by 5/2 to get 0.204 moles.
Now that we know the moles of our products, let's plug some stuff into the ideal gas law PV = nRT. We are looking for P so let's solve for that. P = (nRT)/V, now let's plug in our values. Make sure V is converted to liters so 0.5 L. And convert celcius to kelvin by adding 273
P = ((0.204 moles)(318 K)(0.08206 L atm mol^-1 K^-1))/0.5 L
A lot of units cancel, and we get about 10.65 atm, if you don't want the answer in atm, you can find a conversion equation. But let's round to sig figs for now, which will bring us to 10 atm.
Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol
![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Answer:
Magnesium hydroxide has 3 unique elements.
Explanation:
Long story short, 2 is the balancing compound in structure to make up hydroxide, therefore 1 compound would be left to create Mg(2O)H.
(This is only an opinion of mathematical science to me, I don't have complete understanding of this subject either, good luck.)
