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Inessa05 [86]
3 years ago
9

Thehalf life of carbon is 5730 years. how many years would it take a 200 gram sample to Decay to only 25 G?​

Chemistry
1 answer:
MatroZZZ [7]3 years ago
8 0

When a half-life passes, only half of the mass of the sample stays. So to get to 25 g from 200, let's see how much half-lives will pass.

200÷2=100

100÷2=50

50÷2=25

The answer to that is 3 half-lives, or

3 × 5730 = 17190 years

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A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant K_f = 5.32 de
kirill115 [55]

Answer: -15.4^00C

Explanation:

T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = boiling point of solution = ?

T^o_f = boiling point of solvent (X) = -10.1^oC

k_f = freezing point constant  = 5.32^oC/kgmol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte like urea)

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

M_2 = molar mass of solute (urea) = 60 g/mol

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

(-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}

T_f=-15.4^0C

Therefore, the freezing point of  solution is -15.4^0C

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If my primary DNA strand is ATACCGCAA <br> a write the complimentary DNA strand
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Read 2 more answers
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
3 years ago
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