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Zanzabum
3 years ago
14

Determine the molarity of a solution formed by dissolving 97.7 g libr in enough water to yield 750.0 ml of solution.

Chemistry
2 answers:
Sonja [21]3 years ago
8 0

Hello!

Determine the molarity of a solution formed by dissolving 97.7 g LiBr in enough water to yield 750.0 ml of solution.

We have the following data:  

M (Molarity) =? (in mol / L)

m1 (mass of the solute) = 97.7 g

V (solution volume) = 750 ml → V (solution volume) = 0.75 L

MM (molar mass of LiBr)

Li = 6.941 u

Br = 79.904 u

---------------------------

MM (molar mass of LiBr) = 6.941   + 79.904

MM (molar mass of LiBr) = 86.845 g/mol

Now, let's apply the data to the formula of Molarity, let's see:

M = \dfrac{m_1}{MM*V}

M = \dfrac{97.7}{86.845*0.75}

M = \dfrac{97.7}{65.13375}

M = 1.49999... \to \boxed{\boxed{M \approx 1.5\:mol/L}}\:\:\:\:\:\:\bf\green{\checkmark}

________________________

________________________

*** Another way to solve is to find the number of moles (n1) and soon after finding the molarity (M), let's see:

n_1 = \dfrac{m_1\:(g)}{MM\:(g/mol)}

n_1 = \dfrac{97.7\:\diagup\!\!\!\!\!g}{86.845\:\diagup\!\!\!\!\!g/mol}

n = 1.12499.. \to \boxed{n_1 \approx 1.125\:mol}

M = \dfrac{n_1\:(mol)}{V\:(L)}

M = \dfrac{1.125\:mol}{0.75\:L}

\boxed{\boxed{M = 1.5\:mol/L}}\:\:\:\:\:\:\bf\blue{\checkmark}

_____________________

\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

rewona [7]3 years ago
3 0

Answer:

1.5 M.

Explanation:

  • Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

<em>M = (no. of moles of LiBr)/(Volume of the solution (L).</em>

<em></em>

∵ no. of moles of LiBr = (mass/molar mass) of LiBr = (97.7 g)/(86.845 g/mol) = 1.125 mol.

Volume of the solution = 750.0 mL = 0.75 L.

∴ M = (no. of moles of luminol)/(Volume of the solution (L) = (1.125 mol)/(0.75 L) = 1.5 M.

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gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

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q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

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The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

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The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

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and when the thermal energy  is can be determined by this formula:

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