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2 years ago

What type of bond is formed when an ammonia (NH3) molecule accepts a proton (H+)

Chemistry

2 answers:

mylen [45]2 years ago

8 0

I’m pretty sure it’s NH4, ammonium

kompoz [17]2 years ago

7 0

Answer: coordinate (dative) bond

Explanation:

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Answer pls..... how????

Irina-Kira [14]

Answer:

Explanation:

The definition of words correspond well with the answers I provided and somewhat match with the dictionary definitions.

3 0

2 years ago

HELPP

Natasha_Volkova [10]

I’m pretty sure it’s D.increases the activation energy for a reaction.

4 0

2 years ago

What did Ernest Rutherford s gold foil experiment demonstrate about atoms?

Ghella [55]

-Positively charged nucleus
-Empty spaced
-Dense core

5 0

3 years ago

How many chlorine atoms are on the products side of the reaction 2al 6hcl → 2alcl3 3h2? 2 3 6 9

GenaCL600 [577]

The number of chlorine atoms present on the product side of the reaction is 6

<h3>What is a chemical equation? </h3>

Chemical equations are representations of chemical reactions using symbols and formula of the reactants and products.

The balancing of chemical equations follows the law of conservation of matter which states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.

<h3>How to determine the number of atoms of Cl</h3>

2Al + 6HCl → 2AlCl₃ + 3H₂

Products => AlCl₃ and H₂

Number of Cl atoms = 2 × 3

Number of Cl atoms = 6 atoms

Learn more about chemical equation:

brainly.com/question/7181548

#SPJ4

3 0

2 years ago