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Alexandra [31]
3 years ago
11

What are chemical formulaes? give your own response!!

Chemistry
1 answer:
lbvjy [14]3 years ago
8 0

Answer:

is a way of presenting information about the chemical proportions of atoms

Explanation:

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Determine the empirical formula. a 3.880g sample contains 0.691g of magnesium , 1.84 g of sulfur , and 1.365 g of oxygen .
Aliun [14]

Answer:

Mg S2 O3

Explanation:

.691 g of Mg  is .284 mole

1.84 g of S    is .5739 mole

1.365 g of O is  .8531 mole      you can see the ratio is ~  1 :2 :3

                                                        Mg S2 O3

4 0
2 years ago
Be able to identify the correct elements using clues such as numbers of protons,locations (family/period), number of neutrons, n
Brilliant_brown [7]

an element's name, chemical symbol, atomic number, atomic mass.

IDK what you are even asking for

6 0
3 years ago
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

4 0
3 years ago
What is the definition of a Lewis acid?
Sveta_85 [38]

Answer:

A Lewis acid is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct

Explanation:

CAN YOU MAKE ME BRAINELIST PLEASE

8 0
3 years ago
Bilangan oksidasi vanadium paling tinggi terdapat dalam senyawa..
emmasim [6.3K]
The answer is (b). As, vanadium is attached to five fluoride atoms, each flouride containing -1 oxidation state, hence five fluoride contains -5, to neutralize, vanadium should have +5 oxidation state.

8 0
3 years ago
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