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3 years ago

How many moles of NaCl (s) can be formed from 32 moles of Cl2 (g) reacting with an excess of Na (s)?

Chemistry

1 answer:

blagie [28]3 years ago

7 0

Answer:

64 moles

Explanation:

Let us begin by writing a balanced equation for the reaction. This is illustrated below:

2Na + Cl2 —> 2NaCl

From the equation above,

1 mole of Cl produced 2 moles of NaCl.

Therefore, 32 moles of Cl will produce = 32 x 2 = 64 moles of NaCl.

Therefore, 64 moles of NaCl are produced

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In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

7 0

3 years ago

Read 2 more answers

Why do you think it is important to use the same blanace throughout the entire experiment?

Zepler [3.9K]

It is important to use the same balance throughout the entire experiment since the calibration of each balance is not the same and changing balances could result in a systematic error.

There are three types of errors that could affect the results of the experiment. The effect of random or indeterminate errors is hard to predict, its effect on the results of the experiment could be different every time. The second type of error is the systematic or determinate error, which causes a shift in results in a specific direction. The last type of error in an experiment is human error.

The type of error that could be related to the use of different balances throughout the experiment is the systematic error. Instruments could be a source of error especially if they are poorly calibrated. Also, analytical balances are calibrated differently which may result in inaccuracy in the weighing of chemicals.

To learn more, please refer to brainly.com/question/11541675.

#SPJ4

6 0

2 years ago

If the net force on a object is 375 N to the right, in what direction will the object move?

Viktor [21]

To the right
Because there is an unbalanced force in that direction
:)

8 0

3 years ago

5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H

cricket20 [7]

We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0

3 years ago

Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?

Ber [7]

Answer: C. The Number of Molecules

Explanation: I just took the test, it's correct.

6 0

3 years ago