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3 years ago

13

How many grams are in one mole of CH3CO2H ?

1 answer:

NISA [10]3 years ago

8 0

Answer: 60.2g

Explanation:

One mole acetic acid weighs 60.2g

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What is the total pressure of a gas mixture containing partial pressures of

Advocard [28]

Answer:

1.54 atm

Explanation:

By Dalton's Law Of partial pressure,

Total Pressure = Sum of all partial pressures

So,P= P1 + P2 + P3

Therefore, P=0.23+0.42+0.89

=1.54 atm

8 0

3 years ago

Draw the structural formula of the major product of the reaction of (S)-2,2,3-trimethyloxirane with MeOH, H . Use the wedge/hash

Katarina [22]

Answer:

(S)-3-methoxy-3-methylbutan-2-ol

Explanation:

In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile ( $CH_3OH$ ) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.

See figure 1

I hope it helps!

8 0

4 years ago

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which

Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride = 0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be: $= \dfrac{5 \ kg }{1590 \ kg/m^3}$

= 0.0031 m³

Suppose $\beta$ is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

$\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT$

$In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)$

$V_2 = V_1 \times exp (\beta (T_2-T_1))$

$V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)$

$V_2 = 0.003175 \ m^3$

However; the workdone = -PdV

$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

The heat energy Q = Δ h

$Q = mC_p(T_2-T_1)$

$Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20$

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0

3 years ago

Create a graphic organizer of the water cycle.

aleksandr82 [10.1K]

Answer:

i will do it brainlist plz

Explanation:

5 0

3 years ago

student in chemistry 150-02 weighed out 55.5g of octane c8h18 and allowed it to react with oxygen, o2 the product formed were ca

Tasya [4]

Answer:

C8H8 + 10O2 → 8CO2 + 4H2O

Explanation:

unbalanced reaction:

C8H8 + O2 → CO2 + H2O

balanced for semireactions:

(1) 16H2O + C8H8 → 8CO2 + 40H+

(2) 10(4H+ + O2 → 2H2O)

⇒ 40H+ + 10O2 → 20H2O

(1) + (2):

balanced reaction:

⇒ C8H8 + 10O2 → 8CO2 + 4H2O

8 - C - 8

20 - O2 - 20

8 - H - 8

3 0

4 years ago