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Ivan
2 years ago
5

How long will it take for a 40.0 gram sample of I-131 (half-life = 8.4 days) to decay to 1/16 its original mass?

Chemistry
1 answer:
Shtirlitz [24]2 years ago
6 0

Answer:

42 days

Explanation:

Half life = 8.4 days

Starting mass = 40.0 g

Time = ?

Final Mass = 1/16 * 40 = 2.5 g

First Half life;

Remaining mass = 40 / 2 = 20g

Second Half life;

Remaining mass = 20 / 2 = 10g

Third Half life;

Remaining mass = 10 / 2 = 5g

Fourth Half life;

Remaining mass = 5 / 2 = 2.5g

Time = Number of half lives * Duration of half life = 5 * 8.4 = 42 days

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Blue eyes are recessive. Jonathan is hybrid for blue eyes. His wife Carly, has blue eyes. If they have four children, how many w
astraxan [27]

Answer:

possibly two

Explanation:

you said Jonathan had hybrid eyes, did you mean that he was heterozygous? if so 2 out of 4 will have the recessive gene "blue".

3 0
3 years ago
What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.4
sesenic [268]

Answer: The concentration of Cl^- ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of the CaCl_2

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the AlCl_3

We are given:

n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL  

Putting all the values in above equation, we get

M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M

The concentration of Cl^- ions in the resulting solution will be same as the molarity of solution which is 1.16 M.

Hence, the concentration of Cl^- ions in the resulting solution is 1.16 M.

6 0
3 years ago
A lead–tin alloy of composition 30 wt% Sn–70 wt% Pb is slowly heated from a temperature of 150°C (300°F).(a) At what temperature
Novay_Z [31]

Answer:

a) 231.9 °C

b) 100% Sn

c) 327.5 °C

d) 100% Pb

Explanation:

This is a mixture of two solids with different fusion point:

Tf_{Pb}=327.5 C

Tf_{Sn}=231.9 C

<u>Given that Sn has a lower fusion temperature it will start to melt first at that temperature. </u>

So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.

The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.

At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.

3 0
3 years ago
Will bromine react with sodium and why?
Archy [21]

<span><span>When you write down the electronic configuration of bromine and sodium, you get this

Na:
Br: </span></span>

<span><span />So here we the know the valence electrons for each;</span>

<span><span>Na:  (2e)
Br:  (7e, you don't count for the d orbitals)

Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.

However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>

<span><span /></span><span><span>

</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>

<span>
</span>

4 0
3 years ago
A lab technician mixes a 0.730 M solution of sodium bromide (NaBr) and water. The volume of the solution is 135 milliliters.
Ivenika [448]

Answer:

We need 10.14 grams of sodium bromide to make a 0.730 M solution

Explanation:

Step 1: Data given

Molarity of the sodium bromide (NaBr) = 0.730 M

Volume of the sodium bromide solution = 135 mL = 0.135 L

Molar mass sodium bromide (NaBr) = 102.89 g/mol

Step 2: Calculate moles NaBr

Moles NaBr = Molarity NaBr * volume NaBr

Moles NaBr = 0.730 M * 0.135 L

Moles NaBr = 0.09855 moles

Step 3: Calculate mass of NaBr

Mass NaBr = 0.09855 moles * 102.89 g/mol

Mass NaBr = 10.14 grams

We need 10.14 grams of sodium bromide to make a 0.730 M solution

3 0
3 years ago
Read 2 more answers
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