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34kurt
3 years ago
12

Explain how atomic number relates to the number of particles in an atom’s nucleus.

Chemistry
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

Atoms of each element contain a characteristic number of protons. In fact, the number of protons determines what atom we are looking at (e.g., all atoms with six protons are carbon atoms); the number of protons in an atom is called the atomic number. In contrast, the number of neutrons for a given element can vary.

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Why are the temperatures on other objects in the solar system more extreme than the temperatures on the Earth?
musickatia [10]
Hello there!

It is just based on the fact that they are closer or further away from the Sun.

Hope This Helps You!
Good Luck :) 
7 0
4 years ago
What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)
Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\  ml\\\\V = 33.33 \ ml

Therefore, volume required is 33.33 ml .

5 0
3 years ago
Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c
KiRa [710]

Answer:

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵ . 1/4)

s = 0.016 M

3 0
3 years ago
In a solution with a pH of 3.0, the color of
Alinara [238K]

Answer:

A) litmus is red

Explanation:

To answer this question, it can be helpful to have the color charts. Litmus, phenolphthalein and methyl orange are ways to test the pH of a substance.

<u>Litmus paper</u>

Litmus can tell you if a substance is an acid or a base. You need to put the substance on both red litmus and blue litmus paper.

pH < 7: both papers are red. 3.0 is less than 7.

pH = 7: none of them change color

pH > 7: both papers are blue

<u>Phenolphthalein</u>

When this indicator is added to a substance, the result is either colorless or pink.

0 < pH ≤ 7: colorless. The color is not red or blue for pH 3.0.

pH > 7: pink

<u>Methyl orange</u>

0 < pH < 4: red. The color is not yellow if the pH is 3.0.

4 ≤ pH < 5: orange

pH ≥ 5: yellow

5 0
3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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