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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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Describe the many different forms of energy involved with stretching and releasing a rubber band. What other processes are simil
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Answer:

Conversion of kinetic energy to potential energy (chemo mechanical energy)

In the state of rest, the rubber is a tangled mass of long chained cross-linked polymer that due to their disorderliness are in a state of increased entropy. By pulling on the polymer, the applied kinetic energy stretches the polymer into straight chains, giving them order and reducing their entropy. The stretched rubber then has energy stored in the form of chemo mechanical energy which is a form of potential energy

Conversion of the stored potential energy in the stretched to kinetic energy

By remaining in a stretched condition, the rubber is in a state of high potential energy, when the force holding the rubber in place is removed, due to the laws of thermodynamics, the polymers in the rubber curls back to their state of "random" tangled mass releasing the stored potential energy in the process and doing work such as moving items placed in the rubber's path of motion such as an object that has weight, w then takes up the kinetic energy 1/2×m×v² which can can result in the flight of the object.

Explanation:

5 0
1 year ago
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A driver with a nearly empty fuel tank may say she is "running on fumes."
tensa zangetsu [6.8K]

Answer:

The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.

Explanation:

The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters

We can find the moles of gasoline by using the ideal gas equation: PV = nRT.

Make n (number of moles) the subject of the formula: n = PV/RT, where:

P = 747 mmHg

V = 56.781 liters

R (universal gas constant) = 0.0821 liter·atm/mol·K

T = 25 ∘C = (273 + 25) K = 298 K

1 atm (in the unit of R) = 760 mmHg

Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.

Given that the molar mass of the gasoline = 101 g/mol,

the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g

the density of the liquid gasoline = 0.75 g/mL

In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline

= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons

since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.

6 0
2 years ago
the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many
8_murik_8 [283]

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or  291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

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Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

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100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

1

2

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

1

1

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

1

4

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

6 0
2 years ago
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