Apreschool has campbell's chunky beef soup, which contains 2.5 g of fat and 15 mg of cholesterol per serving (cup), and campbell's chunky sirloin burger soup, which contains 7 g of fat and 15 mg of cholesterol per serving. by combining the soups, it is possible to get 8 servings of soup that will have 29 g of fat and 120 mg of cholesterol. how many cups of each soup should be used?

6 cups of  Campbell's Chunky Beef soup and 2 cups of Campbell's Chunky Sirloin Burger soup should be used.

Explanation:

Let the cups of Campbell's Chunky Beef soup be x

Let the cups of Campbell's Chunky Sirloin Burger soup be y

Total servings or cups required = 8 cups

...(1)

1 cup of Campbell's Chunky Beef soup contains 2.5 g of fat and 15 mg of cholesterol.

Then in x cups of Campbell's Chunky Beef soup have: x(2.5 g+15 mg)

1 cup of Campbell's Chunky Sirloin Burger soup contains 7 g of fat and 15 mg of cholesterol.

Then in y cups of Campbell's Chunky Beef soup have: y(7 g+15 mg)

Total servings or cups must have 29 g of fat and 120 mg of cholesterol.

1 mg = 0.001 g

...(2)

On solving equation (1)& (2) we get;

x = 6 , y = 2

6 cups of  Campbell's Chunky Beef soup and 2 cups of Campbell's Chunky Sirloin Burger soup should be used.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.