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AURORKA [14]
4 years ago
14

How much potassium chlorate (KCIO3) must decompose in order to produce 150 grams of oxygen gas (O2)?

Chemistry
1 answer:
horrorfan [7]4 years ago
4 0

Answer:

765.95 grams.

Explanation:

2 moles of KClO3 produce 3 moles of oxygen.

Using molar masses:

2 * 122.545 g of KCLO3  produces 47.997 grams of oxygen.

47.997 g O2 is produced from 245.09 g KCLO3

150 g    ..........................................    150 * 245.09 / 47.997

= 765.95 grams.

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A tank with a height of 100 feet and a constant cross sectional area of 10 ft has a constant input flow of 15 ft /hour of pulp s
kiruha [24]

Answer:

(a) The consistency as a function of time is C=0.15*t.

(b) The tank will become clogged in 24 minutes.

Explanation:

The rate of accumulation of the pulp stock can be defined as

\frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*C_{o}

In this case, Co is 0, because the exit flow is only water and 0% fiber.

frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*0=Q_{i}*C_{i}

Rearranging adn integrating

dC = (Q_{i}*C_{i})dt\\\int dC = \int (Q_{i}*C_{i})dt\\C=(Q_{i}*C_{i})*t+constant

At t=0, C=0,

C=(Q_{i}*C_{i})*t+constant\\0=(Q_{i}*C_{i})*0+constant\\0=constant\\\\C=(Q_{i}*C_{i})*t

C=(15*0.01)*t=0.15*t

(b) The time at when the concentration reaches 6% is 0.4 hours or 24 minutes.

C=0.15*t\\0.06=0.15*t\\t=0.06/0.15=0.4

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7 0
3 years ago
If a dozen clementine oranges have a mass of 744g, what will be the mass of 15 Clementines?
natka813 [3]
930 g for 15 clementines
5 0
3 years ago
Read 2 more answers
Which of the answer choices best describes how electrons move, according to modern atomic theory?
ivanzaharov [21]
They move in orbitals
7 0
3 years ago
An copper mass is heated and placed in a foam cup calorimeter containing 45.0 mL of water at 24.0 degrees celcius. The water rea
Alenkinab [10]

Answer:

= 9,593.1 Joules

Explanation:

Heat absorbed by water is equivalent to heat released by copper.

Heat absorbed is given by;

Q = mcΔT

where m is the mass, c is the specific capacity and ΔT is the change in temperature.

Therefore;

Since dnsity of water is 1 g/mL, and specific heat capacity is 4.18 J/g°C while the change in temperature is (75-24) = 51°C.

Heat absorbed by water = 45 g × 4.18 J/g°C × 51

                                         = 9,593.1 Joules

Therefore, the heat released by copper is 9,593.1 Joules

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4 years ago
1) Calculate the average speed in meters/minute for the trip represented
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Answer:

in physics we never work with minutes only seconds.

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