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saw5 [17]
3 years ago
12

How do we differentiate chemical change from physical change ​

Chemistry
1 answer:
bija089 [108]3 years ago
5 0

Answer:

Chemical change :has change in mass, heat is needed, new element is formed, hard to reverse.......

Physicalchange:does not have change in mass, heat is not necessary, no new element is formed, easy to reverse

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30 mL of 0.6 M HCl solution is neutralized with 90 mL NaOH solution. What is the concentration of the base? Show all work Please
skad [1K]

From

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<em>n</em><em>b</em><em>=</em><em>1</em>

<em>V</em><em>a</em><em>=</em><em>3</em><em>0</em><em>m</em><em>l</em><em>s</em>

<em>V</em><em>b</em><em>=</em><em>9</em><em>0</em><em>m</em><em>l</em><em>s</em>

<em>M</em><em>a</em><em>=</em><em>0</em><em>.</em><em>6</em><em>M</em>

<em>M</em><em>b</em><em>=</em><em>?</em>

<em>f</em><em>r</em><em>o</em><em>m</em>

<em>Mb</em><em>=</em><em><u>M</u></em><em><u>a</u></em><em><u>×</u></em><em><u>V</u></em><em><u>a</u></em><em><u>×</u></em><em><u>n</u></em><em><u>b</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>V</em><em>b</em><em>×</em><em>n</em><em>a</em>

<em>M</em><em>b</em><em>=</em><em><u>0</u></em><em><u>.</u></em><em><u>6</u></em><em><u>×</u></em><em><u>3</u></em><em><u>0</u></em><em><u>×</u></em><em><u>1</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>9</em><em>0</em><em>×</em><em>1</em>

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<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>9</em><em>0</em>

<em>M</em><em>b</em><em>=</em><em>0</em><em>.</em><em>2</em><em>M</em>

<em>NaOH</em><em>=</em><em>4</em><em>0</em>

<em>F</em><em>r</em><em>o</em><em>m</em>

<em>c</em><em>o</em><em>n</em><em>c</em><em>=</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em>i</em><em>t</em><em>y</em><em>×</em><em>m</em><em>r</em>

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<em>c</em><em>o</em><em>n</em><em>c</em><em>=</em><em>8</em><em>g</em><em>/</em><em>m</em><em>o</em><em>l</em>

3 0
2 years ago
Which statement about atmospheric pressure is false?
Vinil7 [7]

Answer:

D) With an increase in altitude, atmospheric pressure increases as well.

Explanation:

Generally when altitude increases, the value of pressure decreases. This shows that pressure is inversely proportional to altitude. For example, the higher the altitude, the lower the pressure and vice versa. At very high altitude, the number of molecules of air are smaller than the number of moles of air at very low altitude. Thus, the higher the altitude, the lower the atmospheric pressure and the lower the altitude, the higher the atmospheric pressure. Therefore, option (D) is false.

6 0
3 years ago
Read 2 more answers
The compound sodium thiosulfate pentahydrate, Na2S2O3 • 5H2O
Mnenie [13.5K]

The theoretical yield is 204.4 g while the percent yield is 2.57%.

<h3>What is theoretical yield?</h3>

Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.

S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)

Number of moles of sulfur =  3.25 g /8(32) = 0.013 moles

Number of moles of sodium sulfite =  13.1 g/126 g/mol = 0.103 moles

Since 1 moles of sulfur reacts with 8 moles of sodium sulfite

0.013 moles reacts with 0.013 moles ×  8 moles /1 mole = 0.104 moles

There is not enough sodium sulfite hence it is the limiting reactant.

1 mole of sodium sulfite yields 8 moles of product

0.103 moles of sodium sulfite yields  0.103 moles × 8 moles /1 mole = 0.824 moles

Mass of product = 0.824 moles × 248 g/mol = 204.4 g

percent yield =  5.26 g /204.4 g × 100/1

= 2.57%

Learn more about percent yield: brainly.com/question/2506978

4 0
2 years ago
Which opinion would be an example of an alloy <br> brass or <br> silver or <br> gold or <br> copper
frutty [35]

Answer:

Gold and copper are example of alloy

Explanation:

8 0
2 years ago
Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

brainly.com/question/13981382

#SPJ1

7 0
1 year ago
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