You can detect salt in water without tasting by measuring the density of the water. Place a glass of spring water and a glass of the suspected salt water on a balance scale and the heavier one contains salt. Other ways to test for salt in water is to put a drop of water on the end of a nail and place in a gas flame. If the water contains salt, the flame will turn a yellow/orange color.
The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are <span>C) carbon monoxide and carbon dioxide</span>
4 Hydrogen Atoms should be correct. Because the 4 infront of it signifies the amount of hydrogen. It would also be 4 hydrogen atoms if it were written as H4N3, because the 4 is still around the H (as long as the 4 is under scored)
1. 2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. 2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. 2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
<h3>Further explanation</h3>
There are several reactions that can occur in a chemical reaction: single replacement, double replacement, synthesis, decomposition or combustion, etc.
1.Al(s)+HCl(aq)⇒AlCl₃(aq)+H₂(g)
type : single replacement
balance :
2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. AgNO₃ (aq) + Cu (s) ⇒ Cu(NO₃)₂ (aq) + Ag (s)
type : single replacement
balance :
2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. C₃H₈O + O₂ ⇒ CO₂ + H₂O
type : combustion of alcohol
balance :
2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
Answer:
27%
Explanation:
Hello,
The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:
Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:
- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:
As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:
Finally, the percent yield turns out into:
%
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