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Oliga [24]
3 years ago
10

How long will it take for a 750 mg sample of radium with a half life of 15 days to decay to exactly 68mg?

Chemistry
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

52 da  

Step-by-step explanation:

Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.

The i<em>ntegrated rate law for a first-order reaction </em>is  

ln([A₀]/[A] ) = kt

Data:

[A]₀ = 750 mg

 [A] =    68 mg

t_ ½ =   15 da

Step 1. Calculate the value of the rate constant.

 t_½ = ln2/k     Multiply each side by k

kt_½ = ln2         Divide each side by t_½

      k = ln2/t_½

         = ln2/15

         = 0.0462 da⁻¹

Step 2. Calculate the time

ln(750/68) = 0.0462t

         ln11.0 = 0.0462t

            2.40 = 0.0462t     Divide each side by 0.0462

                   t = 52 da

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4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

3 0
2 years ago
A solution containing an equal number of hydrogen ions and hydroxide ions isA. basicB. alkalineC. acidicD. neutral.
Goryan [66]

Answer:

Neutral (Option D)

Explanation:

[H₃O⁺] = [OH⁻]  → Neutral solution   pH = 7

[H₃O⁺] < [OH⁻]  → Alkaline / Basic solution   pH > 7

[H₃O⁺] > [OH⁻] → Acidic solution   pH < 7

5 0
3 years ago
How much water can be raised from 25*C (room temperature) to 37*C (body temperature) by adding the 2,000 kJ in a Snickers bar?
yuradex [85]

Answer:

m = 39834.3 g

Explanation:

Given data:

Mass of water raised = ?

Initial temperature = 25°C

Final temperature = 37°C

Energy added = 2000 Kj (2000 ×1000= 2000,000 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT =  37°C - 25°C

ΔT = 12°C

c = 4.184 g/j.°C

Q = m.c. ΔT

2000,000j = m .4.184 g/j.°C. 12°C

2000,000j = m. 50.208 g/j

m =  2000,000j / 50.208 g/j

m = 39834.3 g

6 0
3 years ago
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