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borishaifa [10]
2 years ago
10

After 17.1 thousand years, what percentage of the original carbon-14 would be left in an organism’s remains? After 17.1 thousand

years, what percentage of the original carbon-14 would be left in an organism’s remains? 25% 17.1% 12.5% 6.25%
Chemistry
1 answer:
hram777 [196]2 years ago
6 0

Answer:

12.5%

Explanation:

Initial percentage of carbon 14 = 100%

Final percentage = ?

Time passed = 17.1 * 1000 years = 17100 years

Half life of carbon 14 = 5,730 years.

So how many Half lives are in 17100 years?

Number of Half lives = Time passed /  Half life = 3.18 ≈ 3

First Half life;

100% --> 50%

Second Half life;

50% --> 25%

Third Half life ;

25% --> 12.5%

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Which is an indicator of poor air quality?<br><br> salt<br> bacteria<br> fertilizer<br> smog
nlexa [21]
Smog

formed by mixture of smoke and fog
3 0
3 years ago
How do you find the molar mass of a hydrate
Darya [45]

If you mean hydrate as in <em>MgSO4 · 7H2O, </em>then simply find the molar mass of each element you see.

For the example above, that means you would add the molar mass (found on the periodic table) of Mg, then S, then 4(O), 14(H), and 7(O).

The results would be your molar mass for the hydrate.

I hope this is what you meant by your question!

6 0
3 years ago
Be sure to answer all parts. From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes non
EleoNora [17]

Answer:

Explanation:

Hrxn = -245.6-296.8+396+50.0 = -96.4 (kJ/mol)

Since Grxn = Hrxn - T*Srxn, where T = 298.15K, we have:

Srxn = (Hrxn - Grxn)/T = (-96.4+75.2)× 1000÷298.15 = -71.1 (j/K mol)

Since: Srxn = Sf (SO2) + Sf (SOCl2)- Sf (SO3) - Sf (SCl2),

or: -71.1 = 248.1 + Sf (SOCl2) - 256.7-184,

Sf (SOCl2) = 256.7+184-71.1-248.1 = 121.5 (J/mol*k)

In order to find T such that Grxn = Hrxn - T*Srxn >= 0. since Hrxn is negative and Srxn is positive, Grxn will always be less than zero. Therefore there won't be a temperature point at which the reaction is going to be non-spontaenous

8 0
3 years ago
What is the ground-state electron configuration of the oxide ion o2−?
bogdanovich [222]

Answer : The ground-state electronic configuration of oxide ion is,

1s^22s^22p^6

Explanation :

Electronic configuration : It is defined as the distribution of electrons of atom in an atomic orbital.

<u>Oxygen</u>

The total number electrons in oxygen = 6

The electronic configuration of oxygen is,

1s^22s^22p^4

<u>Oxide ion</u>

The total number electrons in oxide ion, O^{2-} = 6 + 2 = 8

The charge on O is (-2). So, we are adding 2 electrons.

The electronic configuration of oxide ion is,

1s^22s^22p^6

3 0
3 years ago
The combustion of fuel in your car engine requires oxygen gas, which is supplied as air (21% oxygen molecules) into the engine.
miss Akunina [59]

Answer:

the maximum volume that can be burned per minute is: 0,895 mL of ethanol.

Explanation:

The combustion of ethanol is:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

With gas law:

PV/RT = n

Where P is pressure (1,00 atm)

V is volume (4,73 L of air per minute)

R is gas constant (0,082 atmL/molK)

T is temperatue(25°C≡298,15K)

And n are moles, replacing:

n = 0,193 moles of air per minute.

These moles of air contain:

0,193 moles air ×\frac{21 molesO_2}{100 molesAIR} = <em>0,0406 moles O₂</em>

Thus, the maximum volume that can be burned per minute is:

0,046 moles O₂\frac{1molC_{2}H_{5}OH}{3molesO_2} \frac{46,07 g}{1mol} \frac{1mL}{0,789g} = <em>0,895 mL of ethanol per minute</em>

I hope it helps!

3 0
3 years ago
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