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lora16 [44]
3 years ago
13

PLS HELP ASAP NO LINKS WILL MARK BRAINLIEST IF CORRECT

Chemistry
2 answers:
patriot [66]3 years ago
6 0

Answer:Strontium-90 and cesium-137 have half-lives of about 30 years (half the radioactivity will decay in 30 years). Plutonium-239 has a half-life of 24,000 years. High-level wastes are hazardous because they produce fatal radiation doses during short periods of direct exposure.

Explanation:

Alex777 [14]3 years ago
4 0

Answer:

10,000

Explanation:All toxic waste needs to be dealt with safely, not just radioactive waste. The radioactivity of nuclear waste naturally decays, and has a finite radiotoxic lifetime. Within a period of 10,000 years, the radioactivity of HLW decays to that of the originally mined ore. Its hazard then depends on how concentrated it is.

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What is a temperature scale used by scienctist?
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4 years ago
A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.
Alja [10]

Answer:

  • <u>Question 1: 0.2J/(gºC)</u>
  • <u>Question 2: 6,000J</u>
  • <u>Question 3: 300J</u>
  • <u>Question 4: 80g</u>
  • <u>Question 5: 74ºC</u>
  • <u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

  • Q = m × C × ΔT

<em />

Substitute and solve for C:

  • 566J = 20g × C × (195ºC - 35ºC)
  • C = 566J / (20g × 160ºC)
  • C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

  • Q = m × C × ΔT

<em />

Substitute and compute:

  • Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)
  • Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

  • Q = m × C × ΔT
  • Q = 50g × 0.420J/gºC × (35ºC - 50ºC)
  • Q = 315J

Round to one significant figure (because 50g has one significant figure)

  • Q = 300J

<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

  • Q = m × C × ΔT

Solve for m:

  • m = Q / (C × ΔT)

Substitute and compute:

  • m = 3,000J / (0.712J/gºC × 50ºC)
  • m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. <em>If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.</em>

<em />

  • Q = m × C × ΔT

<em />

Q is negative, since it is released.

Substitute and solve for T:

  • - 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

  • T = - 7500J / 400g × 4.184J/gºC) + 70ºC

  • T = 74ºC

<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

<em />

Question 6. <em>How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC</em>

  • Q = m × C × ΔT

Subsitute:

  • 9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

  • m = 9,500J / (4.184J/gºC × 50ºC)

  • m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

  • m = 50g.
8 0
3 years ago
A compound contains only nitrogen and oxygen and has a formula mass of 44.02. What is the formula of the compound
ch4aika [34]
N₂O

N = 14.01 amu
O ≈ 16 amu

14.01 × 2 = 28.02 ; 2 Nitrogen

44.02 - 28.02 = 16 ; 1 Oxygen

Check

28.02 + 16 ≈ 44.02amu
8 0
3 years ago
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