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3 years ago

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A 1.11 L Flask is filled with 1.15g of argon at 25*C. A Sample of Ethan vapor is added to the same flask until the total pressur

e is 1.450 atm. what is the partial pressure of ethane in the flask?

1 answer:

ycow [4]3 years ago

7 0

Answer:

I thik the pressure will be something about 3 or 4 percent

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A brick measures 25 cm by 12 cm by 13 cm. What is the volume of the brick in cm3? How many milliliters of water would this brick

hodyreva [135]

Volume is lxwxh so do 25x12x13

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3 years ago

Which metalloid is often found in microchips, semiconductors, and sand?a. boron b. arsenic c. germanium d. silicon

elena-14-01-66 [18.8K]

Answer:

silicon

Explanation:

Silicon. Silicon is often found in electronic devices, as well as the sand.

hope this helps UwU

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2 years ago

Read 2 more answers

What is the freezing point (in degrees Celcius) of 4.14 kg of water if it contains 235.1 g of butanol, C 4 H 9 O H

Karolina [17]

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

ΔTf = 8.37 x 3.3585 / 4.14

= 6.79⁰C

Depression in freezing point = 6.79

freezing point of solution = - 6.79⁰C .

5 0

2 years ago

Calculate the mass of 0.25 mole of carbon dioxide

Fantom [35]

Answer:

11 g

Explanation:

1 mole of carbon dioxide =44 g

therefore for 0.25 mole = x

x = 44 × 0.25

x = 11.00g

7 0

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 $C_{0.504}HO_{1.008}$ .

3 0

2 years ago