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3 years ago

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A sample of 25mls of HCL was mixed with 25mls of KOH of the same concentration in a calorimeter. As a result the temperature ros

e from 25'c to 26.6'c. Given that the enthalpy is -56.02kj/mol for the reaction and the specific heat of the dilute aqueous solution is approximately equal to that of water (4.184j/g'C). Determine the molarity of the original HCL solution.

1 answer:

xxTIMURxx [149]3 years ago

4 0

Ooo that cool good thing I learn from this app

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How many moles of carbon tetrachloride are produced when 4.63 mol of carbon reacts with

Ksivusya [100]

Answer:

4.63 moles of CCl4

Explanation:

C + 2Cl2 = CCl4

The balanced equation tells us that we'll obtain 1 mole of CCl4 for every 1 mole of C, assuming we have 2 or more moles of Cl2 (which we do in this problem).

Therefore, if we start with 4.63 moles of C, we'll produce 4.63 moles of CCl4

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3 years ago

30 pts! Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liter

mihalych1998 [28]

First, calculate the number of moles of sodium present with the given mass,

31.5 g of sodium x (1 mol sodium/ 23 g sodium) = 1.37 mol sodium

It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.

mols of H2 = (1.37 mols sodium)(1 mol H2/ 2 mols sodium)
mols of H2 = 0.685 mols H2

Then, at STP, 1 mol of gas = 22.4 L.

volume of H2 = (0.685 mols H2)(22.4 L / 1 mol)
volume of H2 = 15.34 L

Answer: 15.34 L

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4 years ago

What stresses will shift the reaction to increase the products

AveGali [126]

Answer:

Change in molarity, temperature, volume/pressure depending on the conditions given

Explanation:

It really depends on the type of a reaction, however, we may apply general trends and see every possibility:

if we increase the concentration of products, then, according to the principle of Le Chatelier, the equilibrium will shift toward the formation of products;
if we have an endothermic reaction, increasing heat will lead a shift to the right and toward formation of products, since heat might be considered a reactant as well;
if we have an exothermic reaction, removing heat/decreasing temperature will lead to an increase in products, as we're removing one of our products, heat, and system will try to rebuild the amount of heat lost forming the other products as a result as well;
if we have gaseous substances in a reaction, an increase in pressure will shift the equilibrium to the right if we have a greater amount in moles of reactant gases compared to products, this is also known as a decrease in volume;
if we have gaseous substances in a reaction, a decrease in pressure will shift the equilibrium to the right if we have a greater amount in moles of product gases compared to reactants, this is also known as an increase in volume.

7 0

3 years ago

40 g of sodium reacts with ____ g of chlorine to produce 117 g of sodium chloride *

Romashka-Z-Leto [24]

Answer:

9.26g of chlorine

Explanation:

7 0

3 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 + 2\;x & &x&&3-2\;x\end{array}$ .

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

7 0

3 years ago