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3 years ago

How are elements and compounds similar

2 answers:

6 0

I don't like to just give people the answer...so I'll explain what they are as well as differences and you should be able to deduce how they are similar. Compounds are basically a mix of atoms from different elements. An Element is a pure substance made up of one type of atom (you find them on the periodic table). Compounds can be broken down into similar forms...elements cannot. Compounds list is basically endless...so far there are only about 117 known elements. Compounds are expressed as a formula...Elements are expressed using symbols. Hope this helps get you on the right track!

Ludmilka [50]3 years ago

3 0

Elements and compounds are pure chemical substances found in nature. The difference between an element and a compound is that an element is a substance made of same type of atoms, whereas a compound is made of different elements in definite proportions. Examples of elements include iron, copper, hydrogen and oxygen. Examples of compounds include water (H2O) and salt (Sodium Chloride - NaCl)

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2 years ago

Fe2O3+2Al=Al2O3+2Fe

True [87]

We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
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3 years ago

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Answer:

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2 years ago

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At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

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3 years ago

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adelina 88 [10]

Explanation:

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3 years ago