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4 days ago

Reaction pathway A is exothermic. neither exothermic nor endothermic. endothermic. Reaction pathway B is exothermic. neither exo

thermic nor endothermic. endothermic. Which reaction pathway has the faster rate of reaction and why? Reaction pathway B has the faster rate because the products have less energy than the products of reaction pathway A. Reaction pathway A has the faster rate because the products have more energy than the products of reaction pathway B. Reaction pathway B has the faster rate because the activation energy is smaller than the activation energy of reaction pathway A. Reaction pathway A has the faster rate because the activation energy is greater than the activation energy of reaction pathway B. Which reaction pathway has the larger heat of reaction, Δ H rxn ? ΔHrxn?

Chemistry

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Identify the number of moles in 369 grams of calcium hydroxide. Use the periodic table and the polyatomic ion resource.

Alekssandra [3086]

Response: The moles in 369 grams of calcium hydroxide are 4.98 moles

Reasoning: Given,

Mass of calcium hydroxide = 369 g

Molar mass of calcium hydroxide = 74.093 g/mole

Formula used:

$\text{Moles of calcium hydroxide}=\frac{\text{Mass of calcium hydroxide}}{\text{Molar mass of calcium hydroxide}}$

Now substituting the provided values into this formula, you will find the moles of calcium hydroxide.

$\text{Moles of calcium hydroxide}=\frac{369g}{74.093g/mole}=4.98mole$

Thus, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles

7 0

1 month ago

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

lorasvet [2795]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Let's examine the following serial dilution processes.

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is further diluted to 250 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM:

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM:

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is further diluted to 500 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

7 0

2 months ago

An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver c

Tems11 [2777]

Q is determined to be 12.38. The Nernst equation is expressed as Ecell = E°cell - (2.303RT/nF) log Q, where Q represents the reaction quotient. The reaction quotient Q is calculated by taking the product of the products' concentrations divided by the product of the reactants' concentrations. For an electrochemical cell, Q is the concentration ratio of the solution at the anode compared to that at the cathode. Consequently, Q = [anode]/[cathode], specifically Q = 0.052/0.0042, arriving at a value of Q = 12.38.

6 0

2 months ago

In a chemical reaction that takes place at a fixed pressure and volume, the enthalpy change (ΔH) is –585 kJ/mol. Will this react

VMariaS [2998]

According to the sign convention, a negative Δ $H$ indicates that the reaction is exothermic, resulting in a loss of heat and a reduction in temperature.