<u>Answer:</u> The mass of water produced is 507.6 g
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
......(1)
Given mass of glucose = 846 g
Molar mass of glucose = 180 g/mol
Plugging values in equation 1:
![\text{Moles of glucose}=\frac{846g}{180g/mol}=4.7 mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20glucose%7D%3D%5Cfrac%7B846g%7D%7B180g%2Fmol%7D%3D4.7%20mol)
The given chemical equation follows:
![C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O+energy](https://tex.z-dn.net/?f=C_6H_%7B12%7DO_6%2B6O_2%5Crightarrow%206CO_2%2B6H_2O%2Benergy)
By the stoichiometry of the reaction:
If 1 mole of glucose produces 6 moles of water
So, 4.7 moles of glucose will produce =
of water
Molar mass of water = 18 g/mol
Plugging values in equation 1:
![\text{Mass of water}=(28.2mol\times 18g/mol)=507.6g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20water%7D%3D%2828.2mol%5Ctimes%2018g%2Fmol%29%3D507.6g)
Hence, the mass of water produced is 507.6 g
Info: Al(oh)3 might be an improperly capitalized: Al(OH)3
Error: Some elements or groups in the reagents are not present in the products: O
Error: equation Al4C3+H2O=Al(oh)3+CH4 is an impossible reaction
Please correct your reaction or click on one of the suggestions below:
Al4C3 + H2O = Al(OH)3 + CH4
Answer:
<em>0.15M</em>
Explanation:
Molar Concentration = ![\frac{mass}{molar mass}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D)
Assuming 20g of the unadulterated aluminium chloride was weighed into the volumetric flask, and given molar mass of AlCl3 = 133.34g/mol
∴ Molarity = ![\frac{20}{133.34}](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B133.34%7D)
= 0.14999 ≈ <em><u>0.15M (to 2 significant figures)</u></em>
I hope this solution is clear. The same can be calculated from concentration by volume.
Answer:It would be orange
Explanation:I hope this helps
Answer:
![a_x=4.944m/s^2](https://tex.z-dn.net/?f=a_x%3D4.944m%2Fs%5E2)
Explanation:
Hello,
Based on the acting force that is applied horizontally, one can propose the following equation based on Newton's laws:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
Nevertheless, since we've got a angled force, it becomes:
![F=macos\theta](https://tex.z-dn.net/?f=F%3Dmacos%5Ctheta)
In this case,
accounts for the formed angle, so the horizontal acceleration turns out into:
![a_x=\frac{F}{mcos\theta}=\frac{100kg*\frac{m}{s^2} }{25kgcos(36^0)}=4.944m/s^2](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7BF%7D%7Bmcos%5Ctheta%7D%3D%5Cfrac%7B100kg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%7D%7B25kgcos%2836%5E0%29%7D%3D4.944m%2Fs%5E2)
Best regads.