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Degger [83]
1 year ago
12

trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is

still helpful in identifying the presence of the double bond. Select the other signal that would indicate the presence of a C= C bond.
Chemistry
1 answer:
spayn [35]1 year ago
4 0

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 cm^{-1}.

Explanation:

Bands that appear above 3000 cm^{-1}  are often unsaturation diagnoses suggest. The band at 3000- 3100 cm^{-1} is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

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1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore
Alina [70]

Answer :

(1) The theoretical yield of product, MgO is, 1.257 grams.

(2) The percent yield of MgO is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

From the reaction, we conclude that

As, 2 mole of Mg react to give 2 mole of MgO

So, 0.03142 moles of Mg react to give 0.03142 moles of MgO

Now we have to calculate the mass of MgO

\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO

\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g

Theoretical yield of MgO = 1.257 g

Experimental yield of MgO = 0.8922 g

Now we have to calculate the percent yield of MgO

\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100

\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%

Therefore, the percent yield of MgO is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0
2 years ago
The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0
1 year ago
In an experiment, the density of an unknown liquid was calculated to be 0.78 g/ml. if the accepted value is 0.75 g/ml, calculate
Vikki [24]
Percentage error is the relative error your measured value is from the true or accepted value. The formula for percentage error is written below:

Percentage error = |True Value - Measured Value|/True Value   * 100
Percentage error = |0.75 g/mL - 0.78 g/mL|/0.75 g/mL  * 100
Percentage error = 4%
6 0
1 year ago
How many moles of oxygen atoms are in 132.2 g of MgSO4?
zzz [600]

4.4moles of oxygen atoms

Explanation:

Given parameters:

Mass of MgSO₄ = 132.2g

Unknown:

Number of moles of oxygen atoms = ?

Solution:

The number of moles is the quantity of substance that contains the avogadro's number of particles.

 To solve for this;

 Number of moles = \frac{mass}{molar mass}

Molar mass of MgSO₄ = 24 + 32 + 4(16) = 120g/mole

  Number of moles = \frac{132.2}{120} = 1.1 moles

In

     1 moles of MgSO₄ we have 4 moles of oxygen atoms

    1.1 moles of MgSO₄ contains 4 x 1.1 moles = 4.4moles of oxygen atoms

learn more:

number of moles  brainly.com/question/1841136

#learnwithBrainly

8 0
2 years ago
Consider the reaction cacn2 3 h2o → caco3 2 nh3 . how much nh3 is produced if 187 g of caco3 are produced?
shusha [124]
B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.

mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)

mass nh3 = 63.58 g nh3 is produced</span>
7 0
1 year ago
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