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3 years ago

When a metallic element is converted to a positive ion during a chemical reaction with oxygen, the element is said to be?

Chemistry

1 answer:

Orlov [11]3 years ago

5 0

The metal is being <em>oxidized</em>.

Metals- along with other elements- are neutral and have oxidation state of $0$ in their respective elementary states.

Metal atoms can lose electrons to <em>electronegative</em> elements such as oxygen on exposure. Atoms that have lost one or more negatively-charged electrons are no longer neutral and demonstrate a positive charge. Those atoms have converted to positively charged <em>ions</em> or <em>cations</em>.

<em>Oxidation</em> is the process of <em>losing</em> electrons whereas <em>Reduction</em> refers to the process of <em>gaining</em> electrons. "OIL RIG." Metal atoms here lose electrons and are thus oxidized.

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Arrange the following molecules according to the strength of their dispersion forces. Organize in order from largest to smallest

ki77a [65]

Explanation:

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

Therefore, more is the surface area occupied by the carbon chain more will be the dispersion forces present in it. Hence, less is the surface area occupied by a molecule less will be the dispersion forces present in it.

Hence, the given molecules are organized from largest to smallest dispersion forces as follows.

$CH_{3}CH(CH_{3})C(CH_{3})_{2}CH_{2}CH_{3}$ > $CH_{3}CH_{2}(CH_{2})_{4}CH_{2}CH_{3}$ > $CH_{3}CH_{2}CH_{2}CH_{2}CH_{3}$ > $CH_{3}C(CH_{3})2CH_{3}$ > $CH_{3}CH_{2}CH_{3}$ > $CH4$

3 0

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

ad-work [718]

Answer:

-75 cm^3/min

Explanation:

Given from Boyle's law;

PV=C

From product rule;

VdP/dt + PdV/dt = dC/dt

but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min

PdV/dt = dC/dt - VdP/dt

dV/dt = dC/dt - VdP/dt/ P

substituting values;

dV/dt = 0 - (500 * 30)/200

dV/dt = -75 cm^3/min

5 0

3 years ago

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago

If fluorine gas (F₂) occupies a volume of 45.5 L at a pressure of 850

pshichka [43]

Taking into account about the ideal gas law, the mass of fluorine gas is 63.2244 grams.

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

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature. The universal constant of ideal gases R has the same value for all gaseous substances.
R is the ideal gas constant.
n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Mass of fluorine gas</h3>

In this case, you know:

P= 850 mmHg= 1.11842 atm (being 1 mmHg= 0.00131579 atm)
V= 45.5 L
T =100 °C= 373 K (being 0°C= 273 K)
R= 0.082 $\frac{atm L}{mol K}$
n= ?

Replacing in the ideal gas law:

1.11842 atm ×45.5 L = n×0.082 $\frac{atm L}{mol K}$ × 373 K

Solving:

n= (1.11842 atm ×45.5 L)÷ (0.082 $\frac{atm L}{mol K}$ × 373 K)

<u><em>n= 1.6638 moles</em></u>

The molar mass of F₂ is 38 g/mole. Then you can apply the folowing rule of three: If by definition of molar mass 1 mole of the compound contains 38 grams, 1.6638 moles of the compound contains how much mass?

$mass= \frac{1.6638 molesx38 grams}{1 mole}$