You can find moles of a substance by dividing mass by the molecules molar mass.
Explanation:
Carbon Monoxides Molar mass is 12.0111 (Carbon) + 16.0000 (Oxygen) = 28.0111g/mol. Divide grams by molar mass:
<span><span><span>36.55g</span>28.0111</span>=1.305mol<span>s</span></span>
<span><span>
</span></span>
<span><span>
</span></span>
<span><span>
</span></span>
Well. There is the color change and the formation of bubbles.
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
Missing question: What is the vapor pressure of the solution at 25°<span>C?
n(NaCl) = 100 g </span>÷ 58,4 g/mol.
n(NaCl) = 1,71 mol.
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
<span>water's mole fraction = 118,75 mol </span>÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.
Answer:
Letter A
whether a trait is learned or inherited
Explanation:
The nature versus nurture debate involves the extent to which particular aspects of behavior are a product of either inherited (i.e., genetic) or acquired (i.e., learned) influences. Nature is what we think of as pre-wiring and is influenced by genetic inheritance and other biological factors.
