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4 years ago

Determine the molar mass of H2O2 (the solute) in a 1.5M aqueous solution of H2O2

Chemistry

1 answer:

Lera25 [3.4K]4 years ago

3 0

Answer:

The molar mass of H₂O₂ (the solute) in the aqueous solution of

H₂O₂ is 51 g

Explanation:

Given;

H₂O₂ compound

Concentration of aqueous solution of H₂O₂ = 1.5M

The molecular mass of H₂O₂ = (1 x 2) + (16 x 2) = 34 g/mol

$Concentration(M) = \frac{Reacting \ mass \ (g)}{Molar \ mass \ (g/m)}$

Reacting mass (g) = Concentration x Molar mass

Reacting mass (g) = 1.5 x 34

Reacting mass (g) = 51 g

Therefore, the molar mass of H₂O₂ (the solute) in the aqueous solution of

H₂O₂ is 51 g

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The teacher made a 0.5M solution. How is this number read/said?

lukranit [14]

Answer:

It reads as follows: 0.5 moles of solute per liter of solution.

Explanation:

Molarity is the most frequent way of expressing the concentration of solutions in chemistry, and it indicates the number of moles of solute dissolved per liter of solution; is represented by the letter M.

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3 years ago

A mixture of H2 and water vapor is present in a closed vessel at 20°C. The total pressure of the system is 755.0 mmHg.

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Answer:

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4 years ago

What holds the hydrogen and oxegyn atoms together in a water molecule

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Answer: Hydrogen bonds

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3 years ago

The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 271

AfilCa [17]

Answer : The activation energy for the reaction is, 119.7 J

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at 271 K

$K_2$ = rate constant at 281 K = $2K_1$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 271 K

$T_2$ = final temperature = 281 K

Now put all the given values in this formula, we get:

$\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{271K}-\frac{1}{281K}]$

$Ea=119.7J$

Therefore, the activation energy for the reaction is, 119.7 J

3 0

4 years ago

50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol

Viefleur [7K]

Answer:

Approximately $0.47\; \rm mol \cdot L^{-1}$ (note that $1\; \rm M = 1 \; \rm mol \cdot L^{-1}$ .)

Explanation:

The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this $\rm NaNO_3$ solution in water,

Let $n$ be the number of moles of the solute in the whole solution. Let $V$ represent the volume of that solution. The formula for the molarity $c$ of that solution is:

$\displaystyle c = \frac{n}{V}$ .

In this question, the volume of the solution is known to be $1250\; \rm mL$ . That's $1.250\; \rm L$ in standard units. What needs to be found is $n$ , the number of moles of $\rm NaNO_3$ in that solution.

The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of $\rm NaNO_3$ is $85\; \rm g \cdot mol^{-1}$ means that the mass of one mole of

$\displaystyle n = \frac{m}{M}$ .

For this question,

$\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}$ .

Calculate the molarity of this solution:

$\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}$ .

Note that $1\; \rm mol \cdot L^{-1}$ (one mole per liter solution) is the same as $1\; \rm M$ .

8 0

3 years ago