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igomit [66]
3 years ago
5

Cu2(s)+O2(g)=Cu2O(g)+SO2(g)please help urgent will give brainiest​

Chemistry
1 answer:
AysviL [449]3 years ago
3 0

Answer:

2 Cu2S + 3 O2 = 2 Cu2O + 2 SO2

Explanation:

2 Cu2S + 3 O2 = 2 Cu2O + 2 SO2

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In the experiment "Preparation and Standardization of a Sodium Hydroxide Solution", why was the burette rinsed with NaOH before
FromTheMoon [43]

Answer: The correct option is ALL OF THE ABOVE.

Explanation:

TITRATION is a type of volumetric analysis which is used for determining the concentration of solutions. In this process a specific volume of a solution is placed in a conical flask by means of a pipette and small quantities of a second solution is slowly added from a burette until the end point is reached. This is determined by a means of an indicator which shows a characteristic colour change.

During titration, the following precautions should be followed to avoid errors and maintain standardisation in the experiment.

--> Any air bubble in the burette and pipette must be removed during measurement

--> the burette tap should be tightened to avoid leakage.

--> Remove the funnel from the burette before taking any reading to avoid errors in reading the volume.

--> use the base solution such as Sodium Hydroxide Solution to rinse the burette after washing with soap and tap water:

• to remove any air bubble and fill it's tip

• to remove any residual liquid from the water and soap solution which may interfere with the results of the experiment.

• to check if the burette is in good condition.

Therefore all of the above options to the question are correct.

3 0
3 years ago
What is the general trend for valence electrons as you go across the periodic table left to right
Margarita [4]
The atomic radius decreases across a period from left to right and increases down a given group. The atoms with the largest atomic radii are located in group l and at the bottom of groups. Moving from left to right across a period, electrons are added one at a time to the outer energy shell. Hope this helps!
6 0
3 years ago
How many atoms are in 12 moles of silver?
lord [1]

Answer:

NA=72.264x1023

Explanation:

I got you

7 0
3 years ago
A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben
Anit [1.1K]

<u>Answer:</u> The percent yield of the reaction is 32.34 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol

The chemical equation for the reaction of benzoic acid and methanol is:

\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = \frac{1}{1}\times 0.0295=0.108 moles of methyl benzoate

  • Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g

  • To calculate the percentage yield of methyl benzoate, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%

Hence, the percent yield of the reaction is 32.34 %

6 0
3 years ago
How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)
zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0
3 years ago
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