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mr Goodwill [35]
2 years ago
15

At what temperature (Celsius) will water change from a liquid to solid?

Chemistry
2 answers:
Allisa [31]2 years ago
8 0

Answer:

Water changes from a liquid to a solid at 0° Celsius

fiasKO [112]2 years ago
6 0

Answer:

0 degrees Celsius

Explanation:

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1. Which two groups of elements in the periodic table are the most reactive?(1 point) A) alkali metals and halogens B) alkaline
tatyana61 [14]

Answer:

Explanation:

1) alkali metals and halogens as alkalis have 1 electron they need to get rid of in order to gain a full outer shell of electrons and the halogens have to gain 1 electron.

2) they have the same number of electrons on the outer shells

3)they usually have high melting points

4) low or no reactivity

5) group 7

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3 years ago
What is polar and unpolar bonds?
Mars2501 [29]

Answer:Nonpolar bonds form between two atoms that share their electrons equally. Polar bonds form when two bonded atoms share electrons unequally.

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4 years ago
Energy released from atomic reactions is called?
erma4kov [3.2K]
The answer is nuclear energy
7 0
3 years ago
Which model of acids and bases would NOT define NH3 as a base (even though it is considered one by the other models)?
prohojiy [21]

Answer:

<em>The correct option is A) Arrhenius</em>

Explanation:

According to the Arrhenius concept of acids and bases, an acid must produce H+ ions when it is present in a solution and the base must produce OH- ions when placed in a solution.

Ammonia does not contain OH- ions of its own when dissolved in water.

The reaction of ammonia dissolving is water can be written as:

NH3     +    H2O     ⇌   NH4+ + OH−

As we can see from the equation, ammonia does form OH- ions but it does not have OH- ions on its own.

Hence, according to the Arrhenius concept, NH3 is not a base.

4 0
3 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

3 0
3 years ago
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