So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
The effective speed (rms) of the oxygen gas is 293.68 m/s.
<h3>
</h3><h3>What is Root-mean-square velocity?</h3>
Root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules
![v_{rms}=\sqrt[]{\frac{3RT}{M} }](https://tex.z-dn.net/?f=v_%7Brms%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B3RT%7D%7BM%7D%20%7D)
<em>where </em>R = universal gas constant
M = molar mass of the gas in kg/mol
T = temperature in Kelvin
According to the ideal gas law,
PV = nRT
RT = 
Substitute in the rms velocity formula,
![v_{rms} = \sqrt[]{\frac{3PV}{nM} }](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%5B%5D%7B%5Cfrac%7B3PV%7D%7BnM%7D%20%7D)
P = 92 kPa, V = 10 L, n = 2 moles and M = 32 x 10⁻³ kg/mol
![v_{rms} = \sqrt[]{\frac{3\times92\times10}{2\times32\times10^-^3} }](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%5B%5D%7B%5Cfrac%7B3%5Ctimes92%5Ctimes10%7D%7B2%5Ctimes32%5Ctimes10%5E-%5E3%7D%20%7D)
=293.68 m/s
Thus, the effective speed (rms) of O₂ gas is 293.68 m/s.
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Answer:
Answer:
The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which simplifies to 1 : 4.
Explanation:
The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.
The chemical equation given is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.
In this case, they are: 2 mol C₄H₁₀ : 13 mol O₂ : 8 mol CO₂ : 10 mol H₂O
Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:
2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to
1 mol C₄H₁₀ : 4 mol CO₂, or
1 : 2.
Explanation:
