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Assoli18 [71]
2 years ago
14

A student sets up the following equation to convert a measurement.

Chemistry
1 answer:
Andrej [43]2 years ago
7 0

Answer:

<h3>\frac{1,000 g}{1 kg}  ·  \frac{100 cm}{1 m}</h3>

Explanation:

The final answer has a different set of units. In particular, kilograms (kg) changes to grams (g) and meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since both kg and m are located in the numerator, they must be located in the denominators of the conversions.

<u>Proportions:</u>

1 kg = 1,000 g

1 m = 100 cm

<u>The full expression:</u>

<h3> -4.3*10^4 \frac{kg*m}{s}  ·  \frac{1,000 g}{1 kg}  ·  \frac{100 cm}{1 m}  =  ?\frac{g*cm}{s}</h3><h2>                     ^        ^</h2>

As you can see, the old units cancel out and you are left with g and cm in the numerator.

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A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from
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Answer:

The correct answer is: X is nitrogen dioxide, and Y is a metal oxide

Explanation:

Combustion of compound of containing nitrogen and metal will give nitrogen  dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.

The general equation is given as:

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3 years ago
The length of a covalent bond depends upon the size of the atoms and the bond order.
lukranit [14]

Answer:

True

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6 0
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How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
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Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

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If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

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