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o-na [289]
3 years ago
9

I really need help with chemical equations and balanced chemical equations ......

Chemistry
1 answer:
SashulF [63]3 years ago
3 0
What do u need help with? I’m sure I can help you
A chemical equation has 2 sides: the reactives side and the products side
For a chemical equation to be balanced these two side must have the same number of atoms of each type, since in a chemical reaction mass must stay the same

Each symbol in either side will have a subscript. A subscript is the tiny number at the bottom of the symbol. If a symbol has no subscript then that indicate there’s only one atom of that type. However if it has, for example, a 4, then that indicates the are 4 atoms of that type. The thing is these subscripts cannot be changed...
Ex.
HgO —> Hg + O₂
Hg = 1 Hg = 1
O = 1 O = 2
This is an unbalanced equation because the is not oxigen on side that the other.
To balance it u must add a coefficient. This number goes before the whole group of symbols — You can not break apart a group to insert a coefficient in the middle. This number is multiplied by the subscript on each of the symbols in the group. You need to find the correct coefficient to multiply by the elements on both sides of the equation.
Note that this coefficient does no have to be the same on both sides
Answer:
2HgO —> 2Hg + O₂
Hg = 2 Hg = 2
O = 2 O = 2
Explanation:
To fix the oxigen we need to add a coefficient to the side that has less quantity. Because 2 x 1 = 2 (1 Is oxigen’s subscript) we place the 2 in front of the group HgO, But in doing so we damage the mercury, so we need to go to the other side to fix that so we place 2 in front of Hg.

If you need any more help, have some exercises I can further explain or have more doubt about this explanation then let me know and I will help you

Ps: I do recommend you first write the quantity of each atom for each side like I did. It makes it easier to visualize what you have to balance



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\bold{\huge{\orange{\underline{ Solution}}}}

\bold{\underline{ Given :- }}

  • <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
  • <u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

\bold{\underline{ To \: Find :- }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

\bold{\underline{ Let's \:Begin:- }}

<u>We </u><u>know </u><u>that</u><u>, </u>

Amount of heat energy = mass * specific heat * change in temperature

<u>That </u><u>is, </u>

\sf{\red{ Q = mcΔT }}

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

\sf{ Q = 250 × 4.180 ×(0 - 100 )}

\sf{ Q = 250 × 4.180 × - 100 }

\sf{ Q = 250 × - 418}

\sf{\pink{ Q = - 104,500 J }}

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

\bold{\underline{ Now :- }}

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>

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2H2O2 - > 2H2O + O2
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