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steposvetlana [31]
4 years ago
6

Using LeChâtelier’s principle, determine whether the reactants or products are favored and the direction of the shift in equilib

rium (left or right) in the following examples. 2CO(g) + O2(g)  2CO2(g) + heat a. Remove O2 b. Lower the temperature c. Add CO d. Remove CO2 e. Decrease pressure
Chemistry
1 answer:
AnnyKZ [126]4 years ago
6 0

Answer:

a. right

b. right

c. left

d. left

e. left

Explanation:

a. When you remove O2 that means that the reactant side is lacking in an O2 so the equilibrium shifts to the product side.

b. This is an exothermic reaction, because heat is on the products side. So since you are decreasing temperature, then it would shift toward your right.

c. WHen you add CO you are adding more to the reactants side therefore the equilibrium would shift toward the left.

d. When you remove CO2 that means that there is a lack of CO2, therefore the equilibrium would shift to the left.

e. When you decrease pressure, you are increasing in volume. So when you look at volume, if it is increasing you look at larger moles, and if it is decreasing you look at smaller moles. In this case, it is increasing therefore it would shift to the left.

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Answer:

b) \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

c) Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

Explanation:

We assume that part a is test the claim. And we can conduct the following hypothesis test:

Null hypothesis: \mu =7

Alternative hypothesis \mu \neq 7

The statistic is to check this hypothesi is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

We know the following info from the problem:

\bar X = 6.4 , s=0.5, n =30

Replacing we got:

t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573

And the p value would be:

p_v= 2*P(Z

Since the p value is very low compared to the significance assumed of 0.05 we have enough evidence to reject the null hypothesis that the true mean is equal to 7 moles/liter

Part b

The confidence interval is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

Part c

Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

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Consider the unbalanced chemical equation hf + al -> alf3 + h2. when coefficients are added to balance the equation, which tw
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Answer:

a

Explanation:

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What is the molarity of an aqueous solution that contains 78g of C6H12O6 dissolved in 2500 mL of solution?
dusya [7]

Answer:

\boxed {\boxed {\sf molarity = 0.17 \ M \ C_6H_12O_6}}

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

molarity = \frac {moles}{liters}

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.

1. Gram to Moles

We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.

  • C: 12.011 g/mol
  • H: 1.008 g/mol
  • O: 15.999 g/mol

Next, look at the formula and note the subscripts. This tells us the number of atoms in 1 molecule. We multiply the molar mass of each element by its subscript.

6(12.011)+12(1.008)+6(15.999)=180.156 g/mol

Use this number as a ratio.

\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}

Multiply by the given number of grams.

78 \ g \ C_6H_12O_6 *\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}

Flip the fraction and divide.

78 \ g \ C_6H_12O_6 *\frac { 1 \ mol \ C_6H_12O_6}{180.156 \ g\ C_6H_12 O_6}

\frac { 78 \ mol \ C_6H_12O_6}{180.156 }= 0.432958102977 \ mol \ C_6H_12O_6

2. Milliliters to Liters

There are 1000 milliliters in 1 liter.

\frac {1 \ L }{ 1000 \ mL}

Multiply by 2500 mL.

2500 \ mL* \frac {1 \ L }{ 1000 \ mL}

2500 * \frac {1 \ L }{ 1000 }= 2.5 \ L

3. Calculate Molarity

Finally, divide the moles by the liters.

molarity = \frac {0.432958102977 \ mol \ C_6H_12O_6}{ 2.5 \ L}

molarity = 0.173183241191 \ mol \ C_6H_12O_6/L

The original measurement has 2 significant figures, so our answer must have the same. That is the hundredth place and the 3 tells us to leave the 7.

molarity \approx 0.17 \ mol \ C_6H_12O_6 /L

1 mole per liter is also equal to 1 M.

molarity = 0.17 \ M \ C_6H_12O_6

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