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3 years ago

What net force must act on the book to make it accelerate to the left at a rate

Chemistry

1 answer:

Nata [24]3 years ago

3 0

Answer:

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Explanation:

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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 $C_{0.504}HO_{1.008}$ .

3 0

3 years ago

How many mole of fe203 are in 233g og the compound

Vadim26 [7]

The answer to this question
Is : 193g

3 0

3 years ago

What is the difference between parent material

Naily [24]

I don’t understand the question

4 0

4 years ago

Will upvote

nirvana33 [79]

The answer is cold air

3 0

3 years ago

1 C3H8 + 5 O2 --> 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many

UNO [17]

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0

3 years ago