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larisa86 [58]
2 years ago
15

There are two naturally occurring isotopes of indium: indium-113 and indium-115. How many neutrons are in a single atom of indiu

m-115?
49
66
115
164
Chemistry
1 answer:
KonstantinChe [14]2 years ago
5 0
There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66. 
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<span>C4H10 + 6.5 O2 ----> 4CO2 + 5H2O 

2C4H10 + 13 O2 ----> 8CO2 + 10H2O 

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2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O 

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4. Now multiply everything on the equation by two so you have nice integer numbers. 

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the nickel-cadmium battery cell has a standard potential of 1.20 v. the cell reaction is 2 nio(oh)(s) cd(s) 2 h2o(l) → 2 ni(oh)2
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8 0
1 year ago
Metallic bonds form between what kinds of atoms?
valentinak56 [21]

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Explanation:

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3 0
2 years ago
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I need help answering this question
ohaa [14]
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4 0
2 years ago
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The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
2 years ago
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