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Sedbober [7]
3 years ago
9

The goal of this lesson is to answer the lesson question, "what does half-life ‘look like' for a radioactive substance?" fill in

the blanks to complete the hypothesis. hypothesis: if an element is radioactive, then the fraction of radioactive remaining after n half-life cycles should be approximately n because…
Chemistry
2 answers:
Serggg [28]3 years ago
6 0

The half life for radioactive can be calculated as:

N /N0 = (1 /2) ^ n

n = T /T half

According to question there are n number of half life are present which would result in remaining amount of element as n.

disa [49]3 years ago
5 0

Atoms

0.5

There are n number of half life are present which would result in remaining amount of element as n.


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Number of H in 3(NH4)2Cro4
zlopas [31]

The number of H atoms in 3(NH₄)₂CrO₄ = 24

<h3>Further explanation  </h3>

The empirical formula is the smallest comparison of atoms of compound forming elements.  

A molecular formula is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

Subscripts in the chemical formula indicate the number of atoms

The compound of 3(NH₄)₂CrO₄ ( 3 molecules of (NH₄)₂CrO₄ ) :

Number of H :

\tt 4\times 2(subscript)\times 3(coefficient,number~of~molecules)=24~atoms

7 0
2 years ago
Suppose of ammonium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of ammonium c
Marta_Voda [28]

Answer:

Final molarity of ammonium cation in the solution = 0.16 M

Explanation:

Complete Question

Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Solution

2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃

We first convert the given parameters to number of moles

Number of moles = (Mass/Molar mass)

Molar mass of NH₄NO₃ = 80.043 g/mol

Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole

Number of moles = (Concentration in mol/L) × (Volume in L)

Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole

2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄

So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.

So, the number of moles of ammonium ion in the product is also 0.03224 mole.

Molarity = (Number of moles)/(Volume L)

Molarity of ammonium ion = (0.03224/0.2) = 0.1612 mol/L = 0.16 M

Hope this Helps!!!

3 0
2 years ago
When air bags inflate, nitrogen gas is formed from sodium azide, along with solid sodium. what reaction category is this?
adell [148]
The answer is
alkaline silocate
7 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
What was the initial temperature of a gas contained in a 2.0-L bottle at a pressure of 1.00 atm if it was placed under vacuum an
lesya692 [45]
V1/T1 = V2/T2 

Substitute the value use ratio and proportion. Use calculator. 

V1 = (V2 x T1) / T2

1 is initial, 2 is final
4 0
3 years ago
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