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3 years ago

Where is the energy added to a chemical reaction stored

Chemistry

2 answers:

leonid [27]3 years ago

7 0

Answer:

Chemical energy is energy stored in the structure of molecules within the bonds between atoms. Energy can be added or released from a molecule by changing the arrangement of electrons (rearranging chemical bonds).

Explanation:

kirza4 [7]3 years ago

6 0

Chemical energy is stored in the structure of molecules within the bonds between atoms

You might be interested in

Is this statement true or false? Ammonia is a base. true false

Vanyuwa [196]

Answer: ------------------------------------------------------------------------------------------------------------------------

Explanation: True, Ammonia is a base

3 0

2 years ago

C5H6 reacts with itself to form C10H12 according to the following process

swat32

When carbon compounds react with themselves to form a larger molecule the process is called polymerization. Specifically addition polymerization. sometimes molecules containing containing carbon to carbon double bonds can join together to form longer chains. The double bond is broken and the electrons in it join to neighboring molecules.

4 0

3 years ago

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 10.0 gg of aluminum? Express your answ

gayaneshka [121]

Answer:

The answer is 0.370moles (3 s.f)

Explanation:

Step 1: write the balanced equation for the reaction

Equation for the reaction:

2Al(s) + 6HCl(g) ---------> 2AlCl3(g) + 3H2(g)

Step 2: Equate the mole of the needed substances

So therefore,

2moles of aluminum will produce 2 moles of aluminum chloride

Note that:

Relative atomic mass of Al = 27 g/mole

And, 27g of Al = 1 mole of Al.

Step 3: Solve for the required number of mole.

2moles of Al = 2moles of AlCl3

Same as

1 mole of Al = 1 mole of AlCl3

27g of Al = 1 mole of AlCl3

10g of Al = (1/27 * 10g) of AlCl3

10g of Al = 0.370moles of AlCl3

Thanks

4 0

3 years ago

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but

alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

Where A is neutral base and AH⁺ is protonated form

The basic dissociation constant, kb, is:

$K_{b} = \frac{[AH^+][OH^-]}{[A]}$

As pH in stomach is 2,5:

[OH] = $10^{-[14-pH]}$

[OH] = 3,16x10⁻¹² M

Thus:

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$ (1)

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$221359 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$0,0127 = \frac{[AH^+]}{[A]}$

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$314456 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$347851= \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

7 0

3 years ago

The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH):

Nadya [2.5K]

Answer:

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

Explanation:

BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

1 g of NaOH

MW BaCl2 = 137 + (35.5x2) = 208 g

MW NaOH = 23 + 16 + 1 = 40 g

208 g of BaCl2 ------------- 1 mol

1 g of BaCl2 ------------- x

x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

40 g of NaOH ------------ 1 mol

1 g of NaOH ------------ x

x = (1 x 1) / 40

x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

0.0048 mol --------------- x

x = (0.0048 x 1) / 1

x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

171 g of Ba(OH)2 -------------------- 1 mol

x -------------------- 0.0048 mol

x = (0.0048 x 171) / 1

x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

% yield = 0.700 / 0.8208 x 100

% yield = 85.3

Sorry, i don't understand this question

6 0

3 years ago