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Ilya [14]
3 years ago
8

Convert 1.5 km to millimeters, and express the result in scientific notation.

Chemistry
1 answer:
Grace [21]3 years ago
7 0
The answer is 1.5e+6. Hope this helped!
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2. The Density of Water is 1.0000g/cm3. Based on this property of water,
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if the object sank then that object has a greater density then water. if the object floated then its density is lower then water.

Explanation:

lets say object 1 has a density of 24/cm3. the density is greater then water (1.0000g/cm3) so it would sink. now lets say object 2 has a density of 0.79383g/cm3 since it's less then the density of water (1.0000g/cm3) it would float.

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As the [ht] in a solution decreases, what happens to the [OH???
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Why you saying we can download an answer we’re here for an answer
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2 years ago
A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound
Karolina [17]

Answer:

The empirical formula of the compound is C_{0.504}HO_{1.008}.

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen (H) is scaled up to a mole, since it has the molar mass, and both carbon (C) and oxygen (O) are also scaled up in the same magnitude. The empirical formula is of the form:

C_{x}HO_{y}

Where x, y are the number of moles of the carbon and oxygen, respectively.

The scale factor (r), no unit, is calculated by the following formula:

r = \frac{M_{H}}{m_{H}} (1)

Where:

m_{H} - Mass of hydrogen, in grams.

M_{H} - Molar mass of hydrogen, in grams per mole.

If we know that  M_{H} = 1.008\,\frac{g}{mol} and m_{H} = 0.2\,g, then the scale factor is:

r = \frac{1.008}{0.2}

r = 5.04

The molar masses of carbon (M_{C}) and oxygen (M_{O}) are 12.011\,\frac{g}{mol} and 15.999\,\frac{g}{mol}, then, the respective numbers of moles are: (r = 5.04, m_{C} = 1.2\,g, m_{O} = 3.2\,g)

Carbon

n_{C} = \frac{r\cdot m_{C}}{M_{C}} (2)

n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }

n_{C} = 0.504\,moles

Oxygen

n_{O} = \frac{r\cdot m_{O}}{M_{O}} (3)

n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }

n_{O} = 1.008\,moles

Hence, the empirical formula of the compound is C_{0.504}HO_{1.008}.

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