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Alex73 [517]
3 years ago
13

Sheena has a lump of sodium, a lump of potassium and a lump of lithium, but they’ve got mixed up and she doesn’t know which one

happens 1. Which group of periodic table are all these elements in? 2. What id the pattern of reactivity in this group -answer for brainlest
Chemistry
1 answer:
Arada [10]3 years ago
5 0

Answer:

Explanation:

Just saw your request regarding answering this so here it is:

All of them belong of Group 1 in periodic table and thus are highly reactive! Pattern of reactivity for Group 1 (Alkali metals) increases as you move down the group as their radius keeps increasing and thus electrons can be easily lost. Thus, to ID the lumps, Sheena should look at their reactivity and she should get the following trend:

Most reactive: Potassium (K)

Intermediate: Sodium (Na)

Least reactive: Lithium (Li)

Hope it helps!

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Which describes the composition of carbohydrates?
UkoKoshka [18]

Answer: Carbohydrates (carbo- = “carbon”; hydrate = “water”) contain the elements carbon, hydrogen, and oxygen, and only those elements with a few exceptions. The ratio of carbon to hydrogen to oxygen in carbohydrate molecules is 1:2:1.

HOPE THIS HELPS

CAN U GIVE ME BRAINLIEST

3 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)

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