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morpeh [17]
3 years ago
9

Which statement is true for every compound? THIS QUESTION IS WORTH 39 POINTS H.E.L.P

Chemistry
1 answer:
Dima020 [189]3 years ago
3 0

Answer: D.two or more elements are chemically bonded

Explanation:

Compound is a pure substance which is made from atoms of different elements chemically combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. They have different properties as compared to their constituents. Example: water H_2O

2H_2O\rightarrow 2H_2+O_2

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Copper Cu

Thus every compound is made up of two or more elements chemically bonded.

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Can y’all help me with this science question plz
elixir [45]
The tree I know is producer
7 0
3 years ago
When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol
oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, O_2

Solution : Given,

Moles of FeCl_3 = 3.00 mole

Moles of O_2 = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)

From the balanced reaction we conclude that

As, 2 moles of FeCl_3 react with 1 mole of O_2

So, 3.00 moles of FeCl_3 react with \frac{3.00}{2}=1.5 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and FeCl_3 is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, O_2

4 0
3 years ago
How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2
tensa zangetsu [6.8K]
__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂
8 0
3 years ago
What is the volume occupied by 2.50 moles of oxygen gas at STP
oee [108]
1 mole of any gas  under STP has volume 22.4 L
So 2.50 moles of any gas ( including oxygen)
2.50 mol *(22.4L/1 mol)=56.0 L
8 0
3 years ago
Using your knowledge of symbols, determine which compound listed must contain both copper and oxygen.
Colt1911 [192]

Answer:

Malachite

Explanation:

Malachite is the only listed compound that must contain copper and oxygen.

Copper and oxygen are both elements found on the periodic table. They have the following symbols;

         Copper  = Cu

         Oxygen  = O

From the given choices, only option 1 has the symbol Cu and O.

So only malachite contains both copper and oxygen.

4 0
3 years ago
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