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Anuta_ua [19.1K]
3 years ago
6

What is the iupac nomenclature of diethyl methane​

Chemistry
1 answer:
grin007 [14]3 years ago
7 0

Answer:

In chemical nomenclature, the IUPAC nomenclature of organic chemistry is a systematic method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC). ... IUPAC names can sometimes be simpler than older names, as with ethanol, instead of ethyl alcohol.

Explanation:

In chemical nomenclature, the IUPAC nomenclature of organic chemistry is a systematic method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC). ... IUPAC names can sometimes be simpler than older names, as with ethanol, instead of ethyl alcohol.

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A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume
MissTica

Answer:

4.56 X 10^ -4 g/mL

Explanation:

A solution is  prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.

(7.6 X10^-4  gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g

this is dissolved )in 10 m L=45.6 X 10^-4  g/ 10

4.56 X 10^ -4 g/mL

check

6/10 =0.6

4.56/7.6 = o.,6

4 0
3 years ago
Controls are defined as ____​
wariber [46]

Answer:

The second one is the answer

4 0
3 years ago
Read 2 more answers
For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the
ladessa [460]

<span>Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643 1/s.
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

4 0
3 years ago
How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

5 0
3 years ago
5. A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas
marin [14]

Answer:

0.499atm

Explanation:

The formula is

P1/V1 = P2/V2

so:

1.26atm/7.40L = P2/2.93L

then:

(1.26atm/7.40L)*2.93L = P2

= 0.4988918911atm

the answer must have 3 sig figs

8 0
3 years ago
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