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insens350 [35]
3 years ago
13

An organism had 1,000 grams of carbon-14 (a radioactive form of carbon) in it when it died. How much remains after five half-liv

es?
Chemistry
1 answer:
Katarina [22]3 years ago
5 0

Answer:

After 5th half life the remaining mass is 31.25 g.

Explanation:

Given data:

Total mass of carbon-14 = 1000 g

Mass remain after 5 half lives = ?

Solution:

At time zero = 1000 g

At first half life = 1000 g/2 = 500 g

At second half life =  500 g/ 2= 250 g

At third half life = 250 g/ 2 = 125 g

At 4th half life = 125 g/2 = 62.5 g

At 5th half life = 62.5 g/2 = 31.25 g

Thus after 5th half life the remaining mass is 31.25 g.

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Explanation:

Wind helps to get out molecules of water from clothes.

In summer Ball is hard b because due to heat kinetic energy of molecules increases

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Which of the following is a primary cause of LO/TO accidents?
Marianna [84]
<h2>Answer:</h2>

The correct answer is

A) Regular operation

<h2>Explanation:</h2>

Even those workplaces that have established LO/TO processes face challenges, including: Lack of specific procedures written for each piece of equipment identifying all energy sources and energy isolation devices. Lack of comprehensive safety training for everyone in the workplace. Incorrect tag use.

So, regular operation is the primary cause of LO/TO accidents.

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3 years ago
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Need help fast please 15 points!!!
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Answer:

C. disposition and condensation ​

Explanation:

8 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
4 years ago
If an ideal gas has a pressure of 5.49 atm, a temperature of 88.78 °c, and has a volume of 22.03 l, how many moles of gas are in
kirza4 [7]
N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
(R = 22.4/273 = 0.082)
= (5.49 x 22.03)/(0.082 x 361.78) = ?
Put it into the calculator. It's hard to do that on a mobile phone.
7 0
4 years ago
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