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Temka [501]
3 years ago
9

One g of U-235 produces approximately how much more energy than 1 g of coal? (a) 3,000 times, (b) 19,000 times, (c) 80,000 times

,(d) 400,000 times, (e) 2,000,000 times.
Chemistry
1 answer:
Artist 52 [7]3 years ago
3 0

2,000,000 times

Explanation:

One gram of U-235 produces approximately 2,000,000 times more energy than one gram of coal.

The energy in 1g of U-235 is a nuclear energy whereas energy obtained from coal is chemical energy by combustion. During combustion chemical bonds are broken between the carbon atoms.

  • Uranium decays the radioactive fission.
  • When it is bombarded with particles, an artificial disintegration is induced.
  • Most nuclear reactions are chain reactions in which one step leads to another until stability is produced.
  • Each step produces energy for the next process.

Learn more:

Non-renewable resource brainly.com/question/2948717

#learnwithBrainly

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How many grams of h3po4 are in 255 ml of a 4.50 m solution of h3po4?
murzikaleks [220]

H3PO4 has molecular weight of approximately 98 grams per mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times 4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I automatically equate 1 Liter to 1000 mL since the given volume is in mL for easier computation.

7 0
2 years ago
A cube is 4 cm on each side. What is its volume?<br><br> Answer: 64cm^3
Anit [1.1K]

Answer:

yeah

Explanation:

5 0
3 years ago
At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival
evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0
3 years ago
Read 2 more answers
The density of a material is a/an ___________.
Nitella [24]

The density of a material is an intensive property.

<h3>What is intensive property?</h3>

An intensive property of matter is one that does not change with the amount of matter. It is a bulk property, which means that it is a physical property that is independent of sample size or mass. An extensive property, on the other hand, is one that is affected by sample size.

<h3>What factors influence an intensive property?</h3>

Intensive properties are those that are determined solely by the characteristics of the material and not by its quantity - for example, density, temperature, refractive index, color, and pressure. Intensive properties are not additive, which means their value does not change when the amount of material is changed.

Learn more about the intensive property here:-

brainly.com/question/24909279

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5 0
1 year ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

3 0
2 years ago
Read 2 more answers
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