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3 years ago

Wich element is likely to be the most reactive? bromine (Br) chlorine (Cl) fluorine (F) iodine (I)

Chemistry

2 answers:

Simora [160]3 years ago

6 0

Answer:

fluorine (F)

Explanation:

I took the test on edge

wlad13 [49]3 years ago

5 0

most reactive among them is fluorine because it is the most electronegative element

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Based on the molecular structures of chloromethane and methane shown, a student makes the claim that a pure sample of chlorometh

goblinko [34]

Answer:

Yes, chloromethane has stronger intermolecular forces than a pure sample of methane has.

Explanation:

In both methane and chloromethane, there are weak dispersion forces. However, in methane, the dispersion forces are the only intermolecular forces present. Also, the lower molar mass of methane means that it has a lower degree of dispersion forces.

For chloromethane, there is in addition to dispersion forces, dipole-dipole interaction arising from the polar C-Cl bond in the molecule. Also the molar mass of chloromethane is greater than that of methane implying a greater magnitude of dispersion forces in operation.

Therefore, chloromethane has stronger intermolecular forces than a pure sample of methane has.

7 0

3 years ago

How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +

Dominik [7]

1) Write the chemical equation.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: unknown.

3) Moles of CH4.

3.1- Set the equation.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

$(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=$ $n=0.24\text{ }mol\text{ }CH_4$

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

$mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2$

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: unknown.

5.1- Set the equation.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

$(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}$ $V=3.98\text{ }L$

3.98 L of O2 is required to react with 2.0 L CH4.

6 0

2 years ago

A hypothetical wave has 4.6 J of energy. What is its hypothetical, approximate frequency?

Mariulka [41]

Answer:

6.94 × 10^33Hz

Explanation:

E = hf

Where;

E = Energy of wave (J)

h = Planck's constant (6.626 × 10^-34J/s)

f = frequency (Hz)

According to the information provided in this question, the hypothetical energy of the wave is 4.6J

Hence, using E = hf

4.6 = 6.626 × 10^-34 × f

f = 4.6 / 6.626 × 10^-34

f = 0.694 × 10^34

f = 6.94 × 10^33Hz

3 0

3 years ago

PLEASE HELP MEEE I WILL MARK BRAINLIST FOR THIS ITS SUPER SIMPLE....BUT U MUST TYPE EVERYTHING needed ASAP

Tatiana [17]

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It's important to note that human activity has the greatest impact on the amount and quality of wildlife habitat in many areas of the world. Wildlife habitat can be destroyed or its quality diminished as a result of urban growth, agricultural practices, pollution, or habitat destruction.

People can also have a positive impact on wildlife populations through improvement and protection of habitat or ecosystems. Planting trees and shrubs and providing sources of food and water in individual yards is one way people can help wildlife.

5 0

3 years ago

An element has four naturally occurring isotopes with the masses and natural abundances given here. find the atomic mass of the

sergeinik [125]

Abundance of each isotope = 0.1900 x100 = 0.001900
0.2500 = 0.002500
88.43 = 0.8843
11.13 = 0.1113
Atomic mass of the element = (135.90714 x 0.001900) + (137.90599 x 0.002500) + (139.90543 x 0.8843) + (141.90924 x 0.1113)
= 0.258223566 + 0.344764975 + 123.718371749 + 15.794498412
= 140.1 amu is the atomic mass of the element

6 0

3 years ago