What is the mass, in grams, of 0.450 moles of Sb?

Daniel Harris has always enjoyed eating tuna fish. Unfortunately, a study of the mercury content of canned tuna in 2010 found th

meriva

That would be 3 days 6 oz of tuna

4 0

2 years ago

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all

Brilliant_brown [7]

Full Question:

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.

denser than water

burns readily in air

boiling point of –1.1°C

odorless

does not react with water

burns readily in air

does not react with water

<h3>Explanation:</h3>

The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.

Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.

6 0

2 years ago

Determine the number of moles and mass requested for each reaction in Exercise 4.42.

suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

Step 1: Write the balanced equation
Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. 46g of Na react with 1 mol of Cl₂. Since the molar mass of Cl₂ is 71g/mol, then 46g of Na react with 71g of Cl₂.

Step 3:

$10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}$

$10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}$

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

216.5g of HgO form 0.5 moles of O₂. 216.5g of HgO form 16g of O₂.

Step 3:

$1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}$

$1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}$

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

16g of O₂ come from 1 mol of NaNO₃. 16g of O₂ come from 85g of NaNO₃.

Step 3:

$128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}$

$128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}$

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

12 g of C form 1 mol of CO₂. 12 g of C form 44g of CO₂.

Step 3:

$20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2$

$[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2$ [/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

79.5g of CuO come from 1 mol of CuCO₃. 79.5g of CuO come from 123.5g of CuCO₃.

Step 3:

$1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}$

5 0

1 year ago

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Virty [35]

Thee question is incomplete; the complete question is;

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Select one or more:

A: Chlorine

B: Carbon

C: Hydrogen

D: Sulfur

E: Fluorine

F: Oxygen

Answer:

chlorine

sulphur

Explanation:

The octet rule states that, for atoms to be stable, they must have eight electrons on their outermost shells.

This rule is not strictly followed by some elements such as sulphur and chlorine. The atoms of these elements can sometimes expand their octet by utilizing the d-orbitals found in the third principal energy level and beyond.

These leads to formation of compounds in which the central atom has more than eight electrons in its outermost shell.

5 0

2 years ago

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.

The hybridization of each atom is shown in the image.

3 0

1 year ago