A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of

Clarification:

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Answer:

B.  26.0 μL.

Explanation:

Hello,

Considering the provided mass and density, the volume calculates to be:

Thus, the solution is B.  26.0 μL.

Best regards.

Fe 3+ + SCN- --> FeSCN 2+ 

<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>

<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>

Given: Mass of the ionic compound = 10.00 g Mass of water = 75.0 g Initial temperature of water T1= 23.2 C Final temperature of water T2 = 31.8 C Specific heat of water c = 4.18 J/gC To determine: Enthalpy of dissolution of the ionic compound Heat gained by water equation: Q = mcΔT m = mass of water c = specific heat ΔT = change in temperature (T2-T1) Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution) Therefore, q(ionic) = 2696 J ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g Answer: A) enthalpy change = 2.7*10² J/g

The percentage of calcium carbonate that reacted is 2.5%. The reaction in question allows us to determine the equilibrium Kp: Kp = the partial pressure of carbon dioxide, since the other components are solids. We'll apply the ICE table to the provided equilibrium. At the start, we have 0.2 for calcium carbonate with no initial moles of other substances. As the reaction progresses, we set the changes to be -x for calcium carbonate, +x for carbon dioxide, and +x for the other product, leading us to an equilibrium of 0.2-x for calcium carbonate while both other products are at x. Using Kp = Kc(RT)ⁿ, where n represents the mole difference of gaseous products and reactants, we find n to equal 1 for this reaction. With R as the gas constant (8.314 J/mol K) and the temperature at 800 °C (1073 K), we substitute the values accordingly. Upon calculation, we find x = 0.005, which indicates the amount of calcium carbonate that dissociated or reacted, leading us to the reacted percentage.