Explanation:
Cadmium, nickel, chromium, and silver are sometimes used as protective platings. Metals have a wide range of corrosion resistance. The most active metals (i.e., those that tend to lose electrons easily)--such as magnesium and aluminum--corrode easily and are listed at the top of Table 2-1.
Answer:
D) N2O5
Explanation:
The molar mass of a substance is defined as the mass of this substance in 1 mol. To solve this question we must find the molar mass of each option:
<em>Molar mass NO:</em>
1N = 14g/mol*1
1O = 16g/mol*1
14+16 = 30g/mol
<em>Molar mass NO2:</em>
1N = 14g/mol*1
2O = 16g/mol*2
14+32 = 46g/mol
<em>Molar mass N2O:</em>
2N = 14g/mol*2
1O = 16g/mol*1
28+16 = 44g/mol
<em>Molar mass N2O5:</em>
2N = 14g/mol*2
5O = 16g/mol*5
28+80 = 108g/mol
That means the compound with the greatest mass is:
<h3>D) N2O5</h3>
92 i am pretty sure
hope i helped
Explanation:
c I think I am not sure so yh
Answer:
Sr would be the limiting reactant
5 moles
Explanation:
Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.
Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.
Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.
From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.
When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.
For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.