Answer:
The flow rate of a tube is the volume of fluid flowing through the tube per unit time. The flowrate is proportional to the product of the velocity of the fluid through the tube, and the cross-sectional area of the tube.
That is
Q = AV
where
A is the area of the tube
V is the velocity of the tube
The cross-sectional area of the tube is proportional to the radius of the tube. From the above equation, we can deduce that if the velocity of the fluid flowing through the tube is held constant, the flowrate of the fluid through the tube will increase with an increase in the radius of the tube, and it will decrease with a decrease in the radius of the tube.
The percent yield is 71.3 %.
Explanation:
Percent yield is the measure to analyze the success percentage of any experiment .The percent yield of any experiment can be obtained by the ratio of actual or experimental value to expected or theoretical value multiplied with 100.
So, in the present problem, we have obtained 10.7 g of adamantium nitrate from Wolverine's 10 pound claws. So the actual value or the experimental value is the amount of adamantium nitrate obtained from Wolverine's claws.
Thus, the experimental outcome is 10.7 g. While we had expected to recover 15 g of adamantium nitrate. So the theoretical outcome is 15 g.
Thus, the percent yield is 71.3 %.
The answer would for sure be 178 because acid & base make that amount so yea!! I’m talllyyy right
Answer: A thing that died a long time ago, and it's bones were preserved in the ground.
Explanation:
Because yes
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
