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3 years ago

Gaseous chlorine dioxide (ClO2) is used in bleaching flour and municipal water treatment in

Chemistry

1 answer:

trapecia [35]3 years ago

5 0

Taking into account the ideal gas law, the pressure is 2.52 atm.

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas are related by a simple formula called the ideal gas law. This equation relates the three variables if the amount of substance, number of moles n, remains constant. The universal constant of ideal gases R has the same value for all gaseous substances. The numerical value of R will depend on the units in which the other properties are worked.

P×V = n×R×T

In this case, you know:

P=?
V= 500 L
n= 52.1 moles
R= 0.082 $\frac{atmL}{molK}$
T= 22 C= 295 K (being 0 C=273 K)

Replacing in the ideal gas law:

P×500 L = 52.1 moles ×0.082 $\frac{atmL}{molK}$ ×295 K

Solving:

P= (52.1 moles ×0.082 $\frac{atmL}{molK}$ ×295 K)÷ 500 L

Finally, the pressure is 2.52 atm.

Learn more about ideal gas law:

brainly.com/question/4147359?referrer=searchResults

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Can alpha radiation affect electronics

Tom [10]

Answer:

Radiation effects on electrical equipment depend on the equipment and on the type of ionizing radiation to which it is exposed.

First, beta radiation has little, if any, effect on electrical equipment because this type of ionizing radiation is easily shielded. The equipment housing and the construction of the parts within the housing will protect the equipment from beta-radiation (high-energy electrons) exposure.

Gamma radiation is penetrating and can affect most electrical equipment. Simple equipment (like motors, switches, incandescent lights, wiring, and solenoids) is very radiation resistant and may never show any radiation effects, even after a very large radiation exposure. Diodes and computer chips (electronics) are much more sensitive to gamma radiation. To give you a comparison of effects, it takes a radiation dose of about 5 Sv to cause death to most people. Diodes and computer chips will show very little functional detriment up to about 50 to 100 Sv. Also, some electronics can be "hardened" (made to be not affected as much by larger gamma radiation doses) by providing shielding or by selecting radiation-resistant materials.

Some electronics do exhibit a recovery after being exposed to gamma radiation, after the radiation is stopped. But the recovery is hardly ever back to 100% functionality. Also, if the electronics are exposed to gamma radiation while unpowered, the gamma radiation effects are less.

Ionizing radiation breaks down the materials within the electrical equipment. For example, when wiring is exposed to gamma rays, no change is noticed until the wiring is flexed or bent. The wire's insulation becomes brittle and will break and may cause shorts in the equipment. The effect on diodes and computer chips is a bit more complex. The gamma rays disrupt the crystalline nature of the inside of the electronic component. Its function is degraded and then fails as more gamma radiation exposure is received by the electronic component.

Gamma rays do not affect the signals within the device or the signals received by the device. Nonionizing radiation (like radio signals, microwaves, and electromagnetic pulses) DO mess with the signals within and received by the device. I put a cheap electronic game in my microwave oven at home. It arced and sparked and was totally ruined. I didn’t waste any more of my time playing that game.

Hope this helps.

Explanation:

MARK ME AS BARINIEST PLS

5 0

4 years ago

What song gets you in your feels the most? mines what could've been by gone west

IrinaK [193]

Answer:def has to be someone you loved by Lewis Capaldi

Explanation:

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3 years ago

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What do you think of my lab report so far?

8090 [49]

Fix ur transition, it sounds choppy

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3 years ago

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The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r

Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈ 2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0

4 years ago

A chemistry student needs 60.0 g of tetrahydrofuran for an experiment. By consulting the CRC Handbook of Chemistry and Physics,

Evgen [1.6K]

Answer:

67.492cm^3

Explanation:

To get the volume of tetrahydrofuran needed, what the student need to do is to apply mathematical calculations.

We know quite well that to obtain the volume of a substance, given the mass of the substance and the density of the substance is possible.

The volume of the substance is simply the mass of the substance divided by the density of the substance.

This is thus volume = mass/density

The mass given is the question is 60g while the density given is 0.889g/cm^3

The volume = 60/0.889 = 67.492cm^3

8 0

3 years ago