given the balanced equation representing a reaction 2NO+O2->2NO2+energy the mole ratio of NO to NO2 IS

1 answer:

4 0

The problem tells us that the equation is already balanced, so we know that we don't have to change any of the coefficients in front of the compounds. Therefore, we can find the ratio by looking at the coefficients of NO and NO²:

2NO : 2NO²

This can be simplified to 1 : 1. Therefore, it is a 1 to 1 ratio.

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What is the maximum number of grams of copper that could be produced by the reaction of 30.0 of copper oxide with excess methane

Solnce55 [7]

Answer: 24.13 g Cu

Explanation:

<u>Given for this question:</u>

M of CuO = 30 g

m of CuO = 79.5 g/mol

Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol

= 0.38 mol

The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:

CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)

The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side

4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)

From the stoichiometry of the balanced equation:

4 moles of CuO gives 4 moles of Cu

1 mole of CuO gives 1 mol of Cu

0.38 mol of CuO gives 0.38 mol of Cu

Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu

= 0.38 × 63.5 g

= 24.13 grams

Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane

4 0

2 years ago

Which list was easier to sort

drek231 [11]

The answer is "List 2". They're all in scientific notation, so it's easier to organize than List 1, which is in both normal notation and scientific notation.

7 0

3 years ago

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl

nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of caffeine extracted is $P = 39.8 \ mg$