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3 years ago

A solution labeled "0.105 M NaOH" would contain ______________ moles of NaOH in each liter of solution.

Chemistry

1 answer:

dem82 [27]3 years ago

7 0

Answer:

A solution labeled "0.105 M NaOH" would contain 0.105 moles of NaOH in each liter of solution.

Explanation:

The concentration of a solution in Molarity (M) stands for the number of moles of that substance contained in 1 L of solution.

Molarity = Concentration in mol/L = (Number of moles of solute) ÷ (Volume of solution in L)

Molarity = Concentration in mol/L = 0.105 M = 0.105 mol/L

Number of moles of solute = ?

Volume of solution in L = 1 L

0.105 = Number of moles of solute × 1

Number of moles of solute = 0.105 mole

Hence, a solution labeled "0.105 M NaOH" would contain 0.105 moles of NaOH in each liter of solution.

Hope this Helps!!!

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How many moles of phosphoric acid are there in 658 grams of phosphoric acid?

vovangra [49]

<span>The molecular formula for phosphoric acid is H3PO4 and has 97.994 grams per mol. In a sample of 658 grams of phosphoric acid, there are 6.71 mols of phosphoric acid.</span>

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3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago

What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg

IrinaVladis [17]

Answer:

Choice A: Approximately $5.23 \times 10^{-29}$ joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the $\rm 5.81\times 10^{-29}$ kilograms of mass.

$E = m \cdot c^{2}$ ,

where

$E$ stands for energy,
$m$ stands for mass, and
$c$ is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of $c$ ?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, $c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}$ .

Apply the mass-energy equivalence:

$\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}$ .

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

$\rm 1\; J = 1\; N \times 1\; m$ .

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

$\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}$ .