Is LiOH soluble or insoluble?

En el enlace covalente entre azufre y carbono (6 y 4 electrones en su ultima capa respectivamente) la molécula tendrá ?

d1i1m1o1n [39]

Answer:

a- Uno de carbono y dos de azufre

Explanation:

El compuesto formado entre el carbono y el azufre es CS2.

El carbono forma dos enlaces dobles con dos átomos de azufre.

Por lo tanto, el compuesto contiene un átomo de carbono y dos átomos de azufre.

6 0

3 years ago

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a

DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0

3 years ago

Distinguish between atom molecules, and ions

NNADVOKAT [17]

Explanation:

Atoms are the smallest unit of matter that can't be broken down chemically. Molecules are groups of two or more atoms that are chemically bonded. Ions are atoms or molecules that have gained or lost one or more of their valence electrons and therefore have a net positive or negative charge.

7 0

4 years ago

2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??

Zepler [3.9K]

Answer:

$d=1.78\times 10^{-7}\ Dg/\mu m^3$

Explanation:

Given,

The density of Helium is $1.78\times 10^4\ g/cm^3$

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

$d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3$

So, the density of Helium is equal to $1.78\times 10^{-7}\ Dg/\mu m^3$ .

4 0

3 years ago

Which of the following would you except to see in the death of a star that is less than 0.5 solar mass

Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: $1.4 \;M_\odot$ .

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its $0.5 \;M_\odot$ , which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0

3 years ago