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3 years ago

Which of the following is an example of the electrostatic force acting in a atom

2 answers:

7 0

Electrostatic forces work on anything that is charged.
Usually only protons and electrons, but also some other elementary particles.

d. a proton attracting an electron

7nadin3 [17]3 years ago

4 0

Answer:

d. a proton attracting an electron

Explanation:

Electrostatic force is the interaction between bodies that have electrical charges

a and b

The neutrons are particles that have no charge, therefore there are no electrostatic forces acting

Electrons are negatively charged particles and when they interact with each other they repel because they have the same charge

Electrons are particles with negative charge and protons with a positive charge, and when interacting with each other they attract because they have different charge

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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

3 years ago

the failure of a cell to react in a normal matter to insulin is most likely the result of a problem with

aniked [119]

A problem with receptors

8 0

3 years ago

You need to prepare 150 mL of 0.1 M solution of silver chloride. How much silver chloride is required?

Rudiy27

Answer:

amount of silver chloride required is 0.015 moles or 2.1504 g

Explanation:

0.1M AgCL means 0.1mol/dm³ or 0.1mol/L

1L = 1000mL

if 0.1mol of AgCl is contained in 1000mL of solution

then x will be contained in 150mL of solution

cross multiply to find x

x = (0.1*150)/1000

x= 0.015 moles

moles of silver chloride present in 150 mL of solution is 0.15 moles

To convert this to grams, simply multiply this value by the molar mass of silver chloride

molar mass of silver chloride AgCl =107.86 + 35.5

=143.36 g/mol

mass of AgCl = moles *molar mass

=0.015*143.36

=2.1504g

4 0

3 years ago

2SO3(g)⟶2S(s)+3O2(g)ΔH=+790kJ. How many kilojoules are required when 2.1 moles of SO3 reacts?

guapka [62]

Given:

Enthalpy change (ΔH) for SO3 decomposition = +790 kJ

Moles of SO3 = 2.1 moles

To determine:

Energy required when 2.1 moles of SO3 reacts

Explanation:

The decomposition reaction is -

2SO3(g) → 2S(s) + 3O2 (g)

Energy required when 2 moles of SO3 reacts is 790 kJ

Thus, for 2.1 moles of SO3 the energy requirement would be

= 2.1 moles SO3 * 790 kJ/2 moles SO3 = 829.5 kJ

Ans: 830 kJ are required when 2.1 moles of SO3 reacts.

4 0

3 years ago

Read 2 more answers

You are working in your laboratory and decide to do some cleaning. You find a test tube with some brown substance congealed at t

Anna [14]

Answer:

The mystery substance is C. Bromine (Br)

Explanation:

Argon (Ar) is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.

Scandium (Sc) is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.

Both bromine and iodine are halogens (group 17 of the periodic table).

The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.

The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.

Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.

You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.

Therefore, the mystery substance is bromine (Br).

7 0

3 years ago