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VikaD [51]
2 years ago
9

In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so

lution is added from a buret to a colorless, acidified solution of H2O2(aq) in an Erlenmeyer flask. (Note: At the end point of the titration, the solution is a pale pink color.)
At a certain time during the titration, the rate of appearance of O2(g) was 1.0 x 10-3 mol/(L⋅s). What was the rate of disappearance of MnO4- at the same time. (please explain it)

Options
6.0 x 10-3 mol/(L⋅s)

A

4.0 x 10-3 mol/(L⋅s)

B

6.0 x 10-4 mol/(L⋅s)

C

4.0 x 10-4 mol/(L⋅s)
Chemistry
1 answer:
pickupchik [31]2 years ago
8 0

Answer:

C. 4x10⁻⁴ mol / (Ls)

Explanation:

Based in the reaction:

5 H₂O₂(aq) + 2 MnO₄⁻(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 O₂(g)

2 moles of MnO₄⁻ disappears while 5 moles of O₂ appears.

If 5 moles appears in a rate of 1.0x10⁻³mol /(Ls), 2 moles will disappear:

2 moles ₓ (1.0x10⁻³mol /(Ls) / 5 moles) = <em>4x10⁻⁴ mol / (Ls)</em>

Right answer is:

C. 4x10⁻⁴ mol / (Ls)

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Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1
pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0
2 years ago
The molecular weight of a gas is ________ g/mol if 6.7 g of the gas occupies 6.3 l at stp.
PIT_PIT [208]
At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol. 
                               n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
                                          6.7g/0.28125 mol = 23.82 g/mol 
6 0
2 years ago
If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?
blagie [28]

Answer:

The answer to your question is     V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

                V1/T1 = V2/T2

-Solve for V2

                V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

                V2 = (750 x 328) / 298

-Simplification

                V2 = 246000 / 298

-Result

                V2 = 825.5 ml

             

7 0
2 years ago
Which element is used in light bulbs as a filament
mojhsa [17]
<span>The element that is used in light bulbs as a filament is tungsten - this is almost always the case in halogen and incandescent bulbs. Tungsten is chosen for this purpose because of the fact it can withstand temperatures of up to 4,500 degrees, as well as being incredibly flexible.</span>
8 0
2 years ago
A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere
kvv77 [185]
Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m =  the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
I_{g} =  \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} =  \frac{m}{12} \, kg-m^{2}

According to the parallel axis theorem, the moment of inertia about the first axis is
I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} =  \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
\frac{I_{g}}{I_{1}} =  \frac{0.0833m}{0.1233m} =0.6756

Answer:  0.676

6 0
2 years ago
Read 2 more answers
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