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chubhunter [2.5K]
3 years ago
9

What is the chemical name of the covalent compound SF6?

Chemistry
2 answers:
dalvyx [7]3 years ago
8 0

Answer:

monosulfur hexafluoride

Explanation:

In the given covalent compound, SF₆, there is one sulfur atom and six fluorine atoms.

The cationic part is sulfur (less electro-negative) and anionic part is fluorine (more electro-negative).

So the more electro-negative fluorine will be named as fluoride and the name of sulfur will be the same :sulfur:

as there are six fluorine so the prefix " hexa"

Hence name will be monosulfur hexafluoride

abruzzese [7]3 years ago
7 0
The answer to that question is letter A
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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) +
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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: <em>unknown</em>.

3) Moles of CH4.

<em>3.1- Set the equation.</em>

PV=nRT

<em>3.2- Plug in the known values and solve for n (moles).</em>

(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: <em>unknown</em>.

<em>5.1- Set the equation.</em>

PV=nRT

<em>5.2- Plug in the known values and solve for V (liters).</em>

(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<em> is required to react with 2.0 L CH4.</em>

.

6 0
1 year ago
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