Se valoró una alícuota de 10ml de HCl con NaOH AL 0.0548M de la cual se gastó 17.9ml. La concentración molar del ácido es:

What's the answer to this question?

Lena [83]

The answer is B- 124 degrees

4 0

2 years ago

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Using the equation below, if you have 4.3 mol of nitrogen tribromide and

ankoles [38]

Answer:

sodium hydroxide is the limiting reactant

Explanation:

The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.

2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr

Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.

For nitrogen tribromide

From the balanced reaction equation;

2 moles of nitrogen tribromide yields 1 mole of nitrogen gas

4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas

For sodium hydroxide;

3 moles of sodium hydroxide yields 1 mole of nitrogen gas

5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas

Therefore, sodium hydroxide is the limiting reactant.

8 0

2 years ago

The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of ther

timama [110]

Answer:

Silver, 0.239 J/(g °C)

Explanation:

The heat change is related to specific heat as given by the formula;

Heat change = mass of substance × specific heat × change in temperature

Therefore; considering same amount of substance or equal masses and have the same initial temperature.

The change in temperature will be inversely proportional to the specific heat.
Therefore; the higher the specific heat lower the temperature change.
Hence, the change in temperature will be highest for the substance with the lowest specific heat.

Therefore; the one that will increase in temperature the most is Silver

7 0

3 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s