The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.
Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
Convert 55.0g Ca(OH)2 to moles.
The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.
To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.
If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
The amount of heat needed to elevate an object's temperature can be determined using the formula,
heat = mass x specific heat x (T2 - T1)
Thus, specific heat can be found with the following formula,
specific heat = heat / (mass x (T2 - T1))
By substituting,
specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The calculated specific heat from the above equation is 0.838 J/g°C.
Explanation:
It is established that 1 gram is equivalent to 1000 milligrams. We can express this mathematically in the following way.
or 
Thus, to convert grams to milligrams, we simply multiply the number by 1000. Conversely, for converting milligrams back to grams, we divide by 1000.