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Shkiper50 [21]
3 years ago
8

If you start with 500.0 g of a radioisotope with a half-life of 3.0 days, how much of the original isotope will still be in the

sample after 18 days?
Chemistry
2 answers:
fomenos3 years ago
8 0
Half life means it decrease by half every 3 days
250 - 3 days
125 - 6 days
62.5 - 9 days
31.25 - 12 days
15.625 - 15 days
7.8125 - 18 days
7.8125 is your final answer.

Your welcome.
Please mark me the brainliest if your are satisfied. 
ValentinkaMS [17]3 years ago
6 0
After 3 days  = 500g becomes 250g
After 6 days  = 250g becomes 125g
After 9 days  = 125g becomes 62.5g
After 12 days = 62.5g becomes 31.25g
After 15 days = 31.25g becomes 15.625g
After 18 days = 15.625g becomes 7.8125g

After 18days, you would have about 7.813g left.
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What is the formula for the compound by calcium ions and chloride ions?
dimaraw [331]
To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
8 0
3 years ago
In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?
Mashutka [201]
Molar mass :

NaBr = 103 g/mol

Pb(NO3)2 = 331.20 g/mol

<span><span /><span>Balanced chemical equation :

</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
</span><span>
2*103 g NaBr ------------> 1 * 331.20 g Pb(NO3)2
      g NaBr  -------------------> 311 g Pb(NO3)2

331.20  g  =   2*103*311

331.20 g = 64066

mass ( NaBr ) =  64066 / 331.20

mass ( naBr)  = 193,43 g of NaBr

hope this helps!.


</span>
5 0
3 years ago
Convert 55.6 km to m <br> show equation
crimeas [40]

Explanation:

1 km = 1000m

55.6 km = x m

x = 55.6*1000

x = 55600m

5 0
3 years ago
Read 2 more answers
How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?
lara [203]
<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

  • <em>q</em> is heat (in J)
  • <em>m</em> is mass (in g)
  • <em>c</em> is specific heat (in J/g °C)
  • ΔT is change in temperature (in °C or K)

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

  1. Substitute in variables [Specific Heat Formula]:                                          q = (225 g)(4.184 J/g °C)(159.8 °C)
  2. Multiply:                                                                                                           q = (941.4 J/°C)(159.8 °C)
  3. Multiply:                                                                                                           q = 150436 J

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

5 0
3 years ago
How many moles of I2 will form 3.58 g of NI3?
Mila [183]

1.       The balanced chemical reaction is:

 

N2 +3 I2 = 2NI3

 

We are given the amount of product formed. This will be the starting point of our calculations.

 

3.58 g NI3 ( 1 mol NI3 / 394.71 g NI3 ) ( 3 mol I2 / 2 mol NI3 ) = 0.014 mol I2.

 

Thus, 0.014 mol of I2 is needed to form the given amount of NI3.

6 0
3 years ago
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