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4 years ago

How do protons neutrons and electrons affect the overall charge of an atom

Chemistry

1 answer:

aleksklad [387]4 years ago

5 0

Answer:

It determines the stability electric state of the atom

Explanation:

Now atoms are stable depending on the number of neutrons they have in their nuclei. If the no of electrons Equal no of protons then the atom is said to be neutral

bear in mind that neutrons have no charge so they don't affect the atom's electric state.

it's the neutrons that determines the atom's radioactive state

You might be interested in

A chemical reaction that has a positive δg is best described as

Alina [70]

Answer:

Explanation:

endergonic

A chemical reaction that has a positive ΔG is correctly described as A) endergonic.

5 0

4 years ago

Can someone please answer question (ii), I thank.

aleksley [76]

The lime water would become cloudy :)

6 0

4 years ago

Consider the generic acid, HA, and how it interacts with water:HA (aq) + H2O (l) ⇌ H3O+ (aq) + A-Consider the following statemen

luda_lava [24]

Answer:

B. False

A. True

B. False in

Explanation:

From Bronsted-Lowrys definition of Acids and bases, a strong acid is a substance that gives up a proton (to form a weak conjugate base), while a strong base is one that willingly accepts a proton.

Therefore in the reaction,

HA(aq) + H2O (l) ⇌ H3O+ (aq) + A-

The stronger HA is, the weaker the A- and vice versa; the weaker HA is, the stronger A-. Example, HCl is a strong acid and its conjugate base, Cl- is a weak base.

B. False

Electron drawing groups are molecules and/or atoms that enable the release of a proton from a specie. They cause inductive as well as mesomeric effects. Examples, -NO2, -COOH, -OH etc.

Fluoride ion is the most stable in this series because it's the most electronegative while carbon is the least stable because it's the least electronegative. Because of this, we were able to say that H-F was the most acidic, because it had the most stable conjugate base.

A. True.

pKa1 = -log[Ka1]

= -log[1.2 x 10-4]

= 3.92

pKa2 = -log[Ka2]

= -log[1.5 x 10-8]

= 7.82

Using Bronsted-Lowry definition, the smaller the pKa value the more ease the acid loses its proton, that is, the smaller the pKa value, the stronger the acid.

Therefore, pKa2 > pKa1, so Ka value of 1.2 x 10-4 is a stronger acid than Ka value of 1.5 x 10-8

B. False.

7 0

4 years ago

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O)? For example if th

oee [108]

Given :

Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .

To Find :

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .

Solution :

By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

So , volume of solution does not matter .

Moles of oxygen , $n=\dfrac{2.666}{16}=0.167\ mole$ .

Now , molecule of CO contains 1 mole of C .

So , moles of C is also 0.167 mole .

Mass of carbon , $m=12\times 0.167=2\ g$ .

Therefore , mass of carbon is 2 grams .

Hence , this is the required solution .

5 0

3 years ago

14. In the lab, an experimenter mixes 75.0 g of water (initially at 30oC) with 83.8 g of a solid metal (initially at 600oC). At

avanturin [10]

Answer:

Tungsten is used for this experiment

Explanation:

This is a Thermal - equilibrium situation. we can use the equation.

Loss of Heat of the Metal = Gain of Heat by the Water

$-Q_{m}=+Q_{w}\\$

Q = mΔT $C_{p}$

Q = heat

m = mass

ΔT = T₂ - T₁

T₂ = final temperature

T₁ = Initial temperature

Cp = Specific heat capacity

<u>Metal</u>

m = 83.8 g

T₂ = 50⁰C

T₁ = 600⁰C

Cp = $x$

<u>Water</u>

m = 75 g

T₂ = 50⁰C

T₁ = 30⁰C

Cp = 4.184 j.g⁻¹.⁰c⁻¹

$-Q_{m}=+Q_{w}\\$

⇒ - 83.8 x $x$ x (50 - 600) = 75 x 4.184 x (50 - 30)

⇒ $x$ = $\frac{6276}{46090} = 0.13$ j.g⁻¹.⁰c⁻¹

We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹

So metal Tungsten used in this experiment

8 0

4 years ago