C. arteries, veins, capillaries
Given:
257J of heat
5500g of mercury
increase by 5.5
degrees Celsius
Required:
Specific heat of
mercury
Solution:
H
= mCpT
257J = (5500g of
mercury) Cp (5.5 degrees Celsius)
Cp = 8.5 x 10^-3
Joules per gram per degree Celsius
Answer:
I think the answers D. Hope this helps you.
Explanation:
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
To solve this we assume that the hydrogen gas is an
ideal gas. Then, we can use the ideal gas equation which is expressed as PV =
nRT. At a constant pressure and number of moles of the gas the ratio T/V is
equal to some constant. At another set of condition of temperature, the
constant is still the same. Calculations are as follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K
<span>V2 = 12.09 L</span>
Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.
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