B KOH
I would say this is the base for the compound substance
Answer:
Here
Explanation:
This is an acid-base reaction (neutralization): CaCO3 is a base, HCl is an acid.
Answer:<span>d. 145 minutes
</span>
Half-life is the time needed for a radioactive to decay half of its weight. The formula to find the half-life would be:
Nt= N0 (1/2)^ t/h
Nt= the final mass
N0= the initial mass
t= time passed
h= half-life
If 25.0% of the compound decomposes that means the final mass would be 75% of initial mass. Then the half-live for the compound would be:
Nt= N0 (1/2)^ t/h
75%= 100% * (1/2)^ (60min/h)
3/4= 1/2^(60min/h)
log2 3/4 = log2 1/2^(60min/h)
0.41503749928 = -60min/h
h= -60 min / 0.41503749928= 144.6min
Answer:
Y = 62.5%
Explanation:
Hello there!
In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:
![m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2](https://tex.z-dn.net/?f=m_%7BCO_2%7D%5E%7Btheoretical%7D%3D10gCO%2A%5Cfrac%7B1molCO%7D%7B28gCO%7D%2A%5Cfrac%7B2molCO_2%7D%7B2molCO%7D%20%20%2A%5Cfrac%7B44gCO_2%7D%7B1molCO_2%7D%5C%5C%5C%5C%20m_%7BCO_2%7D%5E%7Btheoretical%7D%3D16gCO_2)
Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:
![Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7B10g%7D%7B16g%7D%20%2A100%5C%25%5C%5C%5C%5CY%3D62.5%5C%25)
Best regards!