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Alexeev081 [22]
3 years ago
11

The IUPAC name of the missing product for the following reactions will be. 2-butene + water →

Chemistry
1 answer:
alex41 [277]3 years ago
4 0

Answer:

2-butanol or butan-2-ol

Explanation:

This is an example of hydrolysis of alkene to give alkanol

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Celina has a water sample that’s contaminated with salt and microorganisms. Which method should she use to purify the water?
Alik [6]
B) filtration 
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3 years ago
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In general, what makes one kind of amino acid different from other amino acids?
Vinvika [58]
There is part of an amino acid molecule that is called the R group or side chain.  The side chain of the amino acid called glycine is a single hydrogen atom. The side chain is what differs from amino acid to amino acid.
5 0
3 years ago
Consider the following reaction, which is spontaneous at room temperature. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Is ΔS positive o
Ugo [173]

Answer:

∆H = negative and ∆S = positive.

Explanation:

The reaction given in the question is spontaneous at room temperature ,

hence ,

The the gibbs free energy , i.e. ,∆G will be negative for  spontaneous reaction

According to the formula ,

∆G = ∆H -T∆S

The value of ∆G can be negative , if  ∆H has a negative value and  ∆S has a positive value , because , T∆S  , has a negative sign .

Hence , the answer will be , ∆H = negative and ∆S = positive.

3 0
3 years ago
Given: There are 39.95 grams of Argon (39.95 g/1 mole) and one mole has a volume of 22.4 Liters (1 mole/22.4 L). What is the vol
marta [7]

Answer:

V_2=19.23L

Explanation:

Hello,

In this case, by using the Avogadro's law which allows us to understand the volume-moles behavior as a directly proportional relationship:

\frac{V_2}{n_2} =\frac{V_1}{n_1}

We can compute the volume of 34.3 g of argon by representing it in mole as shown below:

n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol

Thus, we find:

V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L

Best regards.

6 0
3 years ago
Ammonia reacts with sulfuric acid to produce the important fertilizer, ammonium hydrogen sulfate.
Liula [17]

Answer:

404.8g of (NH4)HSO4 is produced.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Temperature (T) = 10°C = 10°C + 273 = 283K

Pressure (P) = 110KPa = 110/101.325 = 1.09atm

Volume (V) = 75L

Step 2:

Determination of the number of mole of ammonia, NH3.

The number of mole (n) of ammonia, NH3 can be obtained by using the ideal gas equation. This is illustrated below:

Note:

Gas constant (R) = 0.0821atm.L/Kmol

Number of mole (n) =?

PV = nRT

1.09 x 75 = n x 0.0821 x 283

Divide both side by 0.0821 x 283

n = (1.09 x 75) /(0.0821 x 283)

n = 3.52 moles

Step 3:

Determination of the number of mole ammonium hydrogen sulfate produced from the reaction.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NH3 + H2SO4 —> (NH4)HSO4

From the balanced equation above,

1 mole of NH3 produced 1 mole of (NH4)HSO4

Therefore, 3.52 moles of NH3 will also produce 3.52 moles of (NH4)HSO4.

Therefore, 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 is produced.

Step 4:

Conversion of 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 to grams. This is illustrated below:

Molar Mass of (NH4)HSO4 = 14 + (4x1) + 1 + 32 + (16x4) = 115g/mol

Number of mole of (NH4)HSO4 = 3.52 moles

Mass of (NH4)HSO4 =..?

Mass = mole x molar Mass

Mass of (NH4)HSO4 = 3.52 x 115 = 404.8g

Therefore, 404.8g of (NH4)HSO4 is produced.

5 0
3 years ago
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