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Brrunno [24]
3 years ago
13

What is the percent of N in Li3N? (Li = 6.94 amu, N = 14.01 amu)

Chemistry
1 answer:
MArishka [77]3 years ago
6 0

Answer:

The percent of n would be 7.10

Explanation:

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Current is applied to a molten mixture of AgF , FeCl2 , and AlBr3 . What is produced at each electrode? STRATEGY Rank the cation
ratelena [41]

Answer:

Cathode: Ag

Anode: Br₂

Explanation:

In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:

Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V

Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V

Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V

As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.

For the anode an oxidation must occurs, so the reactions for the nonmetals are:

F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V

Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V

Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V

For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.

5 0
3 years ago
A + B -&gt;C<br><br><br>Which situation best describes the effect of a limiting reactant?
anygoal [31]

Answer:

The Limiting Reactant is the reactant that when consumed the reaction stops.

Explanation:

7 0
3 years ago
What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?
serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0
3 years ago
What is an individual living entity called?
Citrus2011 [14]

Answer:

it is organism the third one

4 0
3 years ago
Read 2 more answers
The above reaction is run and found to follow second order kinetics with a rate constant of 1.30 x 10-3 M-1sec-1. If the initial
Leya [2.2K]

The final concentration after 172 seconds is 1.15 M. To obtain this answer, use the the integrated rate law for the order of reaction. Based on the problem, this is a second order reaction. The units of the rate constant also support that it is a second order reaction.

Further Explanation:

The solution to this problem is straightforward.

  1. Identify the integrated rate law for the corresponding order of reaction.
  2. Substitute the given values into the integrated rate law equation.
  3. Use algebra to solve for the unknown, C(f).

STEP 1: Since the reaction is a second order reaction, the integrated rate law will be:

\frac{1}{C_{f}} = \ kt \ + \frac{1}{C_{i}}

where:

C(f) is the final concentration

k is the rate constant

t is the time

C(i) is the initial concentration

STEP 2: Plugging in the values given in the problem into the equation, the following equation should be obtained:

\frac{1}{C_{f}} \ = (1.30  \ x \ 10^{-3} \frac{ \ 1 }{ \ M-s})(172 \ s) \ + \ \frac{1}{1.54 \ M} \\

STEP 3: Using algebra to solve for C(f):

\frac{1}{C_{f}} = \ 0.87295\\ \\C_{f} = \ 1.14554 \ M

Since the given values only have 3 significant figures, the final answer must be expressed with 3 significant figures as well.

Therefore,

\boxed {C_{f}\ =  \ 1.15 \ M}

Learn More

  1. Learn more about half life brainly.com/question/4318844
  2. Learn more about first order reaction brainly.com/question/8139015
  3. Learn more about rate law brainly.com/question/1450796

Keywords: rate law, second order, integrated rate law

3 0
3 years ago
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