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3 years ago

The metalloid that has three valence electrons is?

Chemistry

2 answers:

bezimeni [28]3 years ago

8 0

The metalloid that has three valence electrons is Boron~

ad-work [718]3 years ago

4 0

Answer:

The metalloid that has three valence electrons is Boron~

Explanation:

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Through which material do P waves move the fastest.

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C is the correct answer

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inysia [295]

Pressure of the gas inside the container is 662.59 torr.

<h3>What is ideal gas law?</h3>

The ideal gas law (PV = nRT) connects the macroscopic characteristics of ideal gases. An ideal gas is one in which the particles are both non-repellent and non-attractive to one another (have no volume).

The general law of ideal gas can be applied here: PV is equal to nRT, where P is the gas pressure in atm.

V is the number of moles of the gas in a mole, and n is the volume of the gas in L. R is the universal gas constant. T is the temperature(Kelvin) of the gas.

If P and T are different values and n and V are constants, then

(P₁T₂) = (P₂T₁).

P₁ = 735 torr, T₁ = 29°C + 273 = 302 K,

P₂ = ??? torr, T₂ = 62°C + 273 = 335 K.

∴ P₂ = (P₁T₂)/(P₁) = (735 torr)(302 K)/(335 K) = 662.59 torr.

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1 year ago

28. Explain the difference between the processes of melting and freezing?

natali 33 [55]

Freezing something makes it solid and melting something makes it a liquid

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4 years ago

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What happens to isotopes that are unstable?

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<u>Explanation:</u>

Isotopes are defined as the chemical species of the same element which differs in the number of neutrons. The isotopes which are unstable are known as radioactive isotope. A radioactive (unstable )isotope can undergo 3 decay process:

1. Alpha Decay: In this decay process, a larger nuclei decays into smaller nuclei by releasing alpha particle. The particle released has a charge of +2 and a mass of 4 units.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha$

2. Beta-minus decay: In this decay process, a neutron gets converted into a proton and an electron. the particle released during this process is a beta-particle.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

3. Beta-plus decay: In this decay process, a protons gets converted into a neutron and electron-neutrino particle. The particle released during this process is a positron particle.

$_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta$

Isotopes which are unstable in nature can undergo these 3 decay processes.

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3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

The ratio of C : H: O = 1: 1: 1

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

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3 years ago