<span>Dmitri Mendeleev's Periodic Law states that chemical and physical properties repeat themselves in groups of 8.</span>
Answer:
Excess Reagent = oxygen
Explanation:
Limiting reagent: The substance that is totally consumed when the reaction is completed.
Excess reagent: The substance left after the limiting reagent is consumed completely
The balanced chemical equation for formation of water is as follow:
This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.
Hence the ratio in which hydrogen and oxygen gas reacts is 2:1
Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.
or
Here 5 mole of oxygen is reacting but only 2 mole is required .
Oxygen is in excess.
heat is transferred from your hand to the ice cube
Answer:
The correct answer is C. element
Explanation:
The sample cannot be an element because an element - or <em>elemental substance</em> - cannot be decomposed into simpler substances. Thus, it cannot be composed by differents types of atoms. For example, an element is carbon (C).
As the sample contains <u>three types of atoms</u>, it can be a compound, a molecule or a mixture, because they can be composed by different types of atoms - of different chemical elements. For example, the sample could contain the element carbon (C) combined with other elements, for example oxygen (O) or hydrogen (H), amoing others.
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
