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swat32
12 days ago
13

if 2.34 moles of Tin IV Chloride reacts with 4.36 moles of sodium, according to the following reaction, what is the maximum numb

er of moles of tin that can be produced?
Chemistry
1 answer:
eduard [944]12 days ago
7 0

1.09 moles of tin (Sn)

Explanation:

The reaction we are examining involves tin chloride (SnCl₂) reacting with sodium (Na) to yield tin (Sn) and sodium chloride (NaCl):

SnCl₄ + 4 Na → Sn + 4 NaCl

Considering the chemical equation, this results in:

If 1 mole of SnCl₂ reacts with 4 moles of Na,

then 2.34 moles of SnCl₂ will react with X moles of Na.

X = (2.34 × 4) / 1 = 9.36 moles of Na

So, 2.34 moles of SnCl₂ can react with 9.36 moles of Na, but we have only 4.36 moles of Na on hand. Therefore, Na is the limiting reactant. Based on this, we can conclude:

If 4 moles of Na yield 1 mole of Sn,

then 4.36 moles of Na will produce Y moles of Sn.

Y = (4.36 × 1) / 4 = 1.09 moles of Sn

Learn more about:

limiting reactant

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Identify one disadvantage to each of the following models of electron configuration:
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Answer:

Below are the downsides of each electron configuration model:

1). Dot Structures - They consume more space and fail to convey the electron distribution within orbitals.

2). Arrow and line diagrams complicate electron counting and also take up excessive space.

3). Written Configurations do not illustrate electron distribution in orbitals, leading to possible errors in counting electrons.

6 0
3 days ago
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Explain why CaCl2 is likely to have properties similar to those of CaBr2
Alekssandra [962]

Answer:

Both CaCl2 and CaBr2 consist of elements (bromine and chlorine) from the same group (group 7).

Explanation:

In the periodic table, elements are arranged into groups based on their valence electron count in the outermost shell. Elements in the same group, which possess a similar number of valence electrons, typically exhibit similar chemical behaviors.

Chlorine and Bromine in CaCl2 and CaBr2 belong to group 7, known as HALOGENS, characterized by having 7 valence electrons in their outer shell.

The similarity in properties between CaCl2 and CaBr2 arises because both contain Chlorine and Bromine, leading to analogous reactions and behaviors when interacting with other compounds.

5 0
15 days ago
Why the gross reading is needed when doing the titration? ​
Anarel [845]

Answer:

The response is provided below.

Explanation:

Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.

Thus, to mitigate these errors, researchers must engage thoroughly throughout experimentation, and employing gross readings can assist in reducing mistakes when determining the final titre value.

7 0
6 days ago
Which of the options below show typical bonding patterns for a neutral nitrogen atom in a molecule or polyatomic ion?
alisha [964]
More information is needed, but in general, a polyatomic ion consists of multiple atoms bonded together, often with instability that affects their bonding patterns.
7 0
14 days ago
Find the age ttt of a sample, if the total mass of carbon in the sample is mcmcm_c, the activity of the sample is AAA, the curre
castortr0y [918]

Answer:

Explanation:

In a desert cave, an artifact has been discovered. The anthropologists investigating this artifact want to determine its age. They note that the current activity level of the artifact is 9.25 decays/s, and the carbon mass present is 0.100 kg. To ascertain the artifact's age, they will employ specific constants:

r=1.2

The formula for carbon 14 activity is

A=A_0e^{\lambda t}

where,

A_0 is the initial activity of the substance

Now, solve for t

-\lambda t=In\frac{A}{A_0}

t=-\frac{1}{\lambda} In(\frac{A}{A_0} )

=-\frac{1}{\lambda} In(\frac{A}{\lambda r(\frac{m_c}{m_a} )} )

since,

A_0=\lambda r(\frac{m_c}{m_a} )

=-\frac{1}{\lambda} In(\frac{A\ m_a}{\lambda r m_c} )

Thus, the age of the artifact is

=-\frac{1}{\lambda} In(\frac{A\ m_a}{\lambda r m_c} )

=-\frac{1}{1.21\times 10^{-4}} In(\frac{(9.25)(2.32\times 10^{-26}}{1.21\times 10^{-4}(\frac{1}{3.15569\times10^7} )(1.2\times 10^{-12})(0.100)}} )\\\\=6303.4 \ years

to two significant figures = 6300 years

4 0
6 days ago
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