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adell [148]
4 years ago
7

Refer to the following compounds.

Chemistry
1 answer:
goblinko [34]4 years ago
5 0

Answer: The answer is D. This has a Carboxylic Acid group, and is acetic acid, or Ethanoic Acid.

ALWAYS LOOK for the Functional Group in question.

A. Would likely not stay in water, or at least not be acidic, for it is butane gas.

B. Is 1-propanol, and alcohols are not acidic as a rule. Certainly not in water.

C. This is an Ether. It will not give up an H+, it it not an acid.

E. This functional group is an amine, which is more “base” like, since the lone pairs of the Nitrogen atom would tend to attract a H+.

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katrin [286]
The student stirs it because he wants the reaction to take place more quickly.
4 0
3 years ago
3.<br> Which statement(s) is/are TRUE for all chemical reactions?
kenny6666 [7]

Answer:

the statement is or are true for all chemical reaction is false

Explanation:

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3 0
4 years ago
Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3
anyanavicka [17]

Answer:  The percent ionization of CH_3NH_2 in a 0.050 M CH_3NH_2(aq) solution is 8.9 %

Explanation:

CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+

 cM                            0             0

c-c\alpha                       c\alpha            c\alpha

So dissociation constant will be:

K_b=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= concentration = 0.050 M and \alpha = degree of ionisation = ?

K_b=4.4\times 10^{-4}

Putting in the values we get:

4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}

(\alpha)=0.089

percent ionisation =0.089\times 100=8.9\%

8 0
3 years ago
Metallic aluminum reacts with MnO2 at elevated temperatures to form manganese metal and aluminum oxide. A mixture of the two rea
Oksi-84 [34.3K]

Answer:

Explanation:

To find the theoretical yield of the equation. First identify the limiting reactant in a chemical equation.

Step 1: write out the equation and balance it.

Al+ 3mno2=3mn+ 2Alo3.

The limiting reactant is mn02 because it is not found in excess.

Step 2: convert the % to gram . All contain 67.2% mole and mno2 will be 100-67.2= 32.8

All=67.2÷100×290(total gram of the reactants)=194.88g

Mno2=32.8÷100×290g=94.12g.

Step 3:calculate the molar mass of mno2 and that of mn. The atomic mass of mn is 54.9380 and that of oxygen is 16.

Mno2=54.938+ (16)2=86.98g/mol.

Mn=54.938.

Step 4:

From your balanced equation , calculate mn.

94.12g mno2× (1mol mno2÷86.98(molarmass) of mno2×3 mol of mn/4molAl×54.938g of mn÷1mol of mn.

94.12g×1÷86.98g×3÷4×54.938÷1

=44.58g

6 0
3 years ago
What is a common source of background radiation
MAVERICK [17]

The radioactive decay of uranium -238 in the earth's crust produces radioactive  radon gas Rn-222 which escapes into the atmosphere is one common cause of background radiation.

4 0
3 years ago
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