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Reil [10]
3 years ago
15

Sodium-24 has a half life how much sodium-24 will remain in an 18 g sample after 75 hours

Chemistry
1 answer:
frutty [35]3 years ago
3 0

Answer:

\boxed{\text{0.56 g}}

Explanation:

The half-life of Na-24 (15 h) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of                 Fraction          Mass

<u>half-lives   t/da   Remaining   Remaining/g </u>

      0             0           1                      18

       1            15            ½                    9.0

      2           30            ¼                    4.5

      3           45            ⅛                    2.2

      4           60            ⅟₁₆                   1.1

      5           75            ⅟₃₂                   0.56

      6           90           ⅟₆₄                   0.28

We see that  \boxed{\textbf{0.56 g}} remain after five half-lives (75 h).

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Breathing equipment used by rescue workers needs to capture the CO2 the humans breath out and produce O2 for them to breath in,
const2013 [10]

Answer: 4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for reaction of potassium superoxide  with carbon dioxide to produce oxygen and potassium carbonate will be:

4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2

8 0
3 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
3 years ago
How is a mixture such as a plate of pasta different from a mixture such as air
stealth61 [152]

Answer:

Well there is a lot of differences between the two. Its called homogeneous and Heterogeneous mixtures. Homogeneous mixtures are all the substances are evenly distributed throughout the mixture (salt water, air, blood).  Heterogeneous mixtures are the substances that are not evenly distributed (chocolate chip cookies, pizza, rocks). So Pasta with sauce and meatballs is heterogeneous and air is homogeneous




HOPE THIS HELPS HAVE A GREAT DAY!!~

Explanation:

6 0
2 years ago
In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
olga nikolaevna [1]

Answer:

a. The second run will be faster.

d. The second run has twice the surface area.

Explanation:

The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

A=\pi ^{1/3} (6V)^{2/3}

The area of the 10.0 cm³-sphere is:

A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}

The area of each 1.25 cm³-sphere is:

A=\pi ^{1/3} (6. 1.25)^{2/3}=5.61cm^{2}

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²

The ratio of  8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.

6 0
3 years ago
1. Energy changes occur as *
alexandr1967 [171]
Number 3 i think is <span>d.heat moves from an object of higher temperature to an object of lower temperature</span>
3 0
3 years ago
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