Answer:
c. they are all used to diagnose medical conditions
Explanation:
Lactate dehydrogenase, aspartate aminotransfcrase, and creatine kinase all are used to diagnose medical conditions.
Answer:
1. Q = 8.66 KJ
2. Q = 7.58 Kcal
3. Q = 0.71 KJ
4. Q = 24.31 Kcal
Explanation:
1. The quantity of heat absorbed can be determined by:
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g and Δθ is the change in temperature.
= 45.2 × 4.2 × (76.9 - 31.3)
= 8656.704
∴ Q = 8.66 KJ
The quantity of heat absorbed is 8.66 KJ.
2. Q = mcΔθ + mL
Where L is the latent heat of fusion of ice = 334 J.
= m(cΔθ + L)
= 72.1(4.2 × 25.2 + 334)
Q = 31712.464 J
= 7579.466 calories
The total heat released is 7.58 Kcal.
3. Q = mcΔθ
= 55.5 × 0.129 × (123.4 - 24.6)
= 707.3586
The quantity of heat required to increase the temperature of gold is 0.71 KJ.
4. Q = mL
Where: L is the specific latent heat of vaporization = 533 calories.
Q = 45.6 × 533
= 24304.8
The quantity of heat required to change water to steam is 24.31 Kcal.
Answer:
0.2320V
Explanation:
Voltage can be defined as the amount of potential energy available (work to be done) per unit charge, to move charges through a conductor.
Voltage can be generated by means other than rubbing certain types of materials against each other.
Please look at attached file for solution to the problem.
The expected voltage generated by this concentration cell is approximately 0.113 V.
To calculate the voltage generated by the concentration cell, we can use the Nernst equation. The Nernst equation relates the concentration of the ions in the two compartments to the voltage of the cell. The equation is:
E = E° - (RT/nF) ln(Q)
Where:
The reaction quotient (Q) can be calculated using the concentrations of the lead (II) and iodide ions in each compartment.
Since this is a concentration cell, the standard cell potential (E°) for this system is 0 V. Therefore, the equation simplifies to:
E = - (RT/nF) ln(Q)
Now we can calculate the voltage:
The solubility product constant (Ksp) for PbI2 is 1.4 x 10-8. Because PbI2 is in a saturated solution, the concentration of Pb2+ ions and I- ions are both equal to the solubility of PbI2. We can substitute these values into the equation to calculate Q:
Q = [Pb²+] x [I-]
Q = (1.4 x 10-8) x (1.4 x 10-8) = 1.96 x 10-16
Now we can calculate the voltage using the given values:
For the Nernst equation, we need to convert the temperature to Kelvin:
T = 25°C + 273 = 298 K
Substitute the values into the equation:
E = - (8.314 J/mol·K x 298 K / 2 x 96,485 C/mol) ln(1.96 x 10-16)
E ≈ 0.113 V
Therefore, the expected voltage generated by this concentration cell is approximately 0.113 V.
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Answer: pH = 1.19
Explanation:
The formula for Ka is: Ka = [H+][A-]/[HA]
where: [H+] = concentration of H+ ions
[A-] = concentration of conjugate base ions
[HA] = concentration of undissociated acid molecules
Equation of reaction: Cl₂CHCOOH ---> H+ + Cl₂CHCOO-
From the equation above, dichloroacetic acid dissociates one H+ ion for every Cl₂CHCOO- ion,
so [H+] = [Cl₂CHCOO-].
Let x represent the concentration of H+ that dissociates from HA, then [HA] = C - x where C is the initial concentration.
Substituting these values into the Ka equation:
Ka = x · x / (C -x)
Ka = x²/(C - x)
(C - x)Ka = x²
x² = CKa - Kax
x² + Kax - CKa = 0
Solve for x using the quadratic formula:
x = [-b ± √(b² - 4ac)]/2a
Note: There are two solutions for x. However only the positive value of x is used since x represents a concentration of ions in solution, and so cannot be negative.
x = [-Ka + √(Ka² + 4CKa)]/2
Substitute the values for Ka and C in the equation above:
Ka = 5.0 x 10^-2
C = 0.15 M
x = {-5.0 x 10^-2 + √[(5.0. x 10^-2)² + 4(0.15)(5.0 x 10^-2)]}/2
x = (-5.0 x 10^-2 + 1.80 x 10^-1)/2
x = 0.13/2
x = 6.50 x 10^-2
To find pH, we use the formula;
pH = -log[H+]
pH = -log(x)
pH = -log(6.50 x 10^-2)
pH = -(-1.19)
pH = 1.19
liquid 1 and 2 have the same color and mass so the answer would be liquid 1 and 2
Explanation:
hope this is helpful
Iodine value is a measure of the degree of unsaturation in fats and oils. It is essentially the number of grams of iodine consumed by 100 g of fat. If the iodine number is in the range of 0-70 then it is a fat, any value above 70 is considered an oil.
Formula:
Iodine number = (ml of 0.1 N Thiosulphate blank- ml of 0.1N thiosulphate test) * 12.7 *100/1000* wt of sample
vol of thiosulphate required to titrate test sample (given oil) = 1 ml
wt of sample = 0.2 g
Information on the volume of thiosulphate required to titrate the blank solution is essential for calculation.
Iodine number = (X-1.0) * 12.7 * 100/1000* 0.2 = (X-1.0)*6.35
B. 0.0860
C. 0.113
D. 0.761
The dry solute has a mass of 0.086 g. 0.113 g of water evaporated while heating, the concentration in G solute / mL solvent at 30.1 C is 0.88.
To find the concentration in g solute / mL solvent, we first need to find the mass of the solvent left after evaporation:
Mass of water before evaporation = 0.113 g
Mass of water after evaporation = 0.113 g - 0.086 g = 0.027 g
Since 1 g of water has a volume of 1 mL, the volume of water after evaporation is also 0.027 mL.
Next, we need to convert the temperature to Kelvin:
T = 30.1 + 273.15 = 303.25 K
We can now use the formula:
concentration = (mass of solute / mass of solvent) / (1 - (mass of water evaporated / mass of solvent))
Plugging in the values we get:
concentration = (0.086 g / (0.027 g)) / (1 - (0.113 g / (0.086 g + 0.027 g)))
concentration = 0.88 g/mL
Therefore, the answer is A. 0.88.
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Answer:
A
Explanation:
worked for me on acellus