Limes have a [H3O+] of 1.3 x 10-2 mol/L. Their pOH is

1 answer:

4 0

To determine the pOH assuming water is the universal solvent take the value of 10 ^ -14 and then divide it by the hydronium concentration and then take the negative logarithm of the final answer that is the solution to the hydroxide ion concentration in the solution.

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When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac

STatiana [176]

Answer:

$\Delta _{comb}H=-2,093\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

$Q_{rxn}+Q_{cal}=0$

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

$Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ$

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

$n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}$

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1 year ago

Which sentences describe alpha particles? Check all that apply 1. Alpha particles consist of two protons and two neutrons 2. Alp

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Which chemical activates the transformation of trypsinogen to trypsin?

jeka57 [31]

The answer should be <span>enteropeptidase

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A sample of a certain material has a mass of 2.03 × 10–3 g. Calculate the volume of the sample, given that the density is 9.133

liq [111]

2.22x10^-3 would be the answer to the question
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You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$