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ikadub [295]
3 years ago
14

Please help! ill give medal

Chemistry
1 answer:
weeeeeb [17]3 years ago
3 0
An example of a general formula of an acid is
<span>HCl

This is called as hydrochloric acid or hydrogen chloride.

H represents for the atom of Hydrogen
and Cl represents for the atom of Chlorine

Since their charges are -1, and +1, it's ratio is 1:1</span>
You might be interested in
What mass of potassium chloride, KCl, is produced when 12.6 g of oxygen, 02, is produced?
blagie [28]

Mass of KCl= 19.57 g

<h3>Further explanation</h3>

Given

12.6 g of Oxygen

Required

mass of KCl

Solution

Reaction

2KClO3 ⇒ 2KCl + 3O2

mol O2 :

= mass : MW

= 12.6 : 32 g/mol

= 0.39375

From the equation, mol KCl :

= 2/3 x mol O2

= 2/3 x 0.39375

=0.2625

Mass KCl :

= mol x MW

= 0.2625 x 74,5513 g/mol

= 19.57 g

4 0
2 years ago
Can someone please help me!!!
LUCKY_DIMON [66]

sorry but I don't know so sorry

6 0
3 years ago
How many molecules of carbon dioxide are in 9.080 x 10 ^-1 mol?
kakasveta [241]

Answer:

5.46 8 x 10²³ molecules.

Explanation:

  • <em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:</u></em>

1.0 mole → 6.022 x 10²³ molecules.

9.08 x 10⁻¹ mole → ??? molecules.

∴ The no. of molecules of CO₂ are in 9.08 x 10⁻¹ mol = (6.022 x 10²³ molecules) ( 9.08 x 10⁻¹ mole) / (1.0 mol) = 5.46 8 x 10²³ molecules.

6 0
3 years ago
So, I have missed school around three days out of the week. and in science, we are going over the periodic table, I suppose.. an
Sonbull [250]
I am unable to see the attachment pm me and i can help you

3 0
3 years ago
A process at constant T and P can be described as spontaneous if ΔG &lt; 0 and nonspontaneous if ΔG &gt; 0. Over what range of t
creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

                              2 SO3(g)   →          2 SO2(g) + O2(g)

ΔHf° (kJ mol-1)  -395.7                        -296.8

S° (J K-1 mol-1)  256.8                         248.2              205.1

ΔH° =  kJ

S° =  J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine  ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where  ν  is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS =  ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) =  2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ =  197.8 - T(0.1879)

and the reaction will become spontaneous when the term  T(0.1879)  becomes greater that 197.8,

0 = 197.8 - 0.1879 T  ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0
3 years ago
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