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wolverine [178]
2 years ago
12

A rock contains uranium 238 which has a half life of 4.5 billion years. if the rock is crushed and heated, the half life in urai

nium-238 will
Chemistry
1 answer:
labwork [276]2 years ago
3 0
The answer is remain the same.If the rock is cut, crushed, or heated, the length of the half life of the uranium would remain the same since these factors have no effect on half-life.Even though chemical changes were affected by temperature, the physical state, pressure, concentration, the chemical environment within the molecule, and any other changing factors, the half-life of a radioactive substance is independent of these influences.
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If your using Plato the correct answer is (A. It accepts protons)
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3 years ago
A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield  Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

                          x = 8161.24/159.8

                          x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

                59 g of Ni ---------------- 159.8 g of Br₂

                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

                 = 54 g

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Identify the reagents represented by the letters in the above reaction scheme. (enter your answer as a string of letters without
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Please show the whole question including the diagram so that you can receive help with the question.
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Imagine a chemist is in the lab and trying to make some chemical reactions happen. In one reaction she reacts chemicals in an ex
VikaD [51]

Imagine a chemist is in the lab and trying to make some chemical reactions happen. In one reaction she reacts chemicals in an exothermic reaction and there is an increase in entropy. A second chemical reaction she is trying to run is endothermic and there is a decrease in entropy. Which of the two reactions is more likely to occur and why?

4 0
3 years ago
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
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