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gladu [14]
3 years ago
13

Student mixes 5 g of sand into a beaker of 20 g of water. Is this a physical or chemical change-why?

Chemistry
1 answer:
jek_recluse [69]3 years ago
6 0
Physical because it doesn't do with fire
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Kernels of corn are heated and become popcorn. Why is this a chemical change?
goldfiish [28.3K]

Answer:

When popcorn is popped, liquid inside the kernel is changed to steam. Pressure from the steam builds up inside the kernel. When the pressure reached a critical stage the kernel pops turning itself inside out. This is a physical change.

Explanation:

6 0
3 years ago
Read 2 more answers
Why is air considered a mixture not a compound?
blondinia [14]
<span>The best reason I can think of for why we believe that air is a mixture is that freezing air slowly yields different liquids at different temperatures. Liquid nitrogen has a different boiling point than liquid oxygen. They also freeze at different temperatures. If air were only 1 compound, then air in its entirety would have a single boiling point and a single freezing point. </span>
8 0
3 years ago
Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air
Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s

b) for completion of 32% of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-32}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}

t=44.3s

c) for completion of 81 % of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-81}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}

t=191s

4 0
3 years ago
0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) +
Veseljchak [2.6K]

Answer:

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄ ----------------  6 moles of water

                        0.55 moles of H₂SO₄ -----------    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

5 0
3 years ago
Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.
-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (NH_2) and a <em>carboxylic acid</em> group (COOH).  Additionally, we have an <u>alcohol </u>(CH_3CH_2OH) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (OH^-).

When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (H_2O). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base (OH^-) removes the hydrogen from the C=O bond to produce ethyl L-valinate

See figure 1

I hope it helps!

7 0
3 years ago
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