Answer:
The answer to your question is The honda civic hybrid weights 13181.8 N or 1318.2 kg
Explanation:
Data
Weight = 2900 lb
Weight = ? N
mass = ? kg
1 kg ------- 10 N
0.22lb ---- 1N
Use proportions to solve this problem
0.22 lb ------------------- 1 N
2900 lb ----------------- x
x = (2900 x 1) / 0.22
x = 2900 / 0.22
x = 13181.8 N
1 kg ----------------------- 10 N
x ----------------------- 13181.8 N
x = (13181.8 x 1)/10
x = 1318.2 kg
Sorry, but where are the ‘items’?
The pressure of the gas is obtained as 48 atm.
<h3>What is the total pressure?</h3>
Now we know that;
Number of moles of CH4 = 48.0 grams /16 g/mol = 3 moles
Number of moles of H2 = 56.0 grams/2 g/mol = 28 moles
Total number of moles present = 3 moles + 28 moles = 31 moles
Using;
PV =nRT
P = total pressure
V = total volume
n = total number of moles
R = gas constant
T = temperature
P = nRT/V
P = 31 * 0.082 * 286/15
P = 48 atm
Learn more about pressure of a gas:brainly.com/question/18124975
#SPJ1
Answer:
I think the answer is 4) 41
Explanation:
APE= atomic number, proton and the electrons are the same number
MAN= mass = atomic number - neutrons
121 - 80 = 41
i haven't done this in a while so hope this helps :)
Answer:
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Explanation:
Hello,
In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:
![[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D2.0x10%5E%7B-3%7D%5Cfrac%7BmolCa%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B2molF%5E-%7D%7B1molCa%5E%7B2%2B%7D%7D%20%20%5C%5C)
![[F^-]_{max}=4x10{-3}\frac{molF^-}{L}](https://tex.z-dn.net/?f=%5BF%5E-%5D_%7Bmax%7D%3D4x10%7B-3%7D%5Cfrac%7BmolF%5E-%7D%7BL%7D)
Best regards.