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2 years ago

What would be the predominant state of matter if there were no intermolecular forces?

1 answer:

4 0

Stronger Intermolecular forces (greater attractions) hold molecules together better. Meaning the molecules don't want to leave whatever liquid they are part of. for example, chloroform(the knock out gas), which has only weak intermolecular forces evaporates very quickly at room temp. However, water which participates in H-bonding, strong intermolecular forces, evaporates slowly at room temp. The colder a liquid is, the harder it is for the molecules to overcome their intermolecular bonds and break free. a.k.a. slow evaporation. but when a liquid is heated the molecules have enough energy to overcome their brotherly bonds and go into the gas phase.

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!!29 POINTS!! PLS HELP ASAP

ICE Princess25 [194]

Answer:

The person who just wanted points is annoying

Explanation:

WHY WOULD YOU DO THAT

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2 years ago

Chemists use activity series to determine whether which type of reaction will occur?

ruslelena [56]

Single replacement reactions. For example copper is more reactive than silver. So a copper wire in a silver solution will cause the silver to become a metal again.

5 0

3 years ago

Read 2 more answers

Based on the activity series, which metals could X represent in the reaction below? (Note: The equation is not balanced.)

Nikitich [7]

Answer is: A) Sr (strontium).
The reactivity series is a series of metals from highest to lowest reactivity. Metal higher in the reactivity series will displace another.
Strontium is only higher in this group from magnesium. Strontium is stronger reducing agent than magnesium, gives electrons easier.

3 0

3 years ago

You have a solution that is 0.02M formate (HCOO-) and 0.03M formic acid (HCOOH), which has a Ka of 1.8x10-4. What is the pH of t

lisov135 [29]

Answer:

3.6

Explanation:

Step 1: Given data

Concentration of formic acid: 0.03 M

Concentration of formate ion: 0.02 M

Acid dissociation constant (Ka): 1.8 × 10⁻⁴

Step 2: Calculate the pH

We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the Henderson-Hasselbach equation.

$pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6$

6 0

2 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

2 years ago