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3 years ago

Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm

Chemistry

1 answer:

pav-90 [236]3 years ago

7 0

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

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What kind of product is formed when a ketone reacts with a grignard reagent followed by protonation?

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c. a tertiary alcohol; when a ketone reacts with a grignard reagent followed by protonation a tertiary alcohol is formed.

More about tertiary alcohol:

No hydrogen atoms are bonded to the functional group's carbon in a tertiary alcohol. Alcohols that have a hydroxyl group bonded to the carbon atom and are linked to three alkyl groups are referred to as tertiary alcohols. These alcohols' structural makeup largely determines their physical characteristics.

This -OH group's existence enables alcohols to create hydrogen bonds with the atoms next to them. Because of this weak connection, alcohols have higher boiling points than their alkane counterparts.

The alcohol is referred to as a tertiary (3°) alcohol if the carbon atom carrying the alcohol group is connected to three other carbon atoms in the alcohol molecule.

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2 years ago

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3 years ago

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

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3 years ago

Chlorophyll pigments are separated using paper. what is the stationary phase in this experiment

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Paper is a stationary phase

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What is the temperature of a water body containing a 90% saturation and 11.5 ppm of dissolved oxygen?

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Using the chart that has been provided, we may determine water temperature. We do this by drawing a straight line form the bottom scale which has the ppm of oxygen dissolved to the middle scale which has the percentage saturation.

The line starts from 11.5 ppm on the bottom scale and goes to 90% on the middle scale. Next, we continue this line, without changing its slope, to the third scale showing temperature. We see that it crosses the temperature scale at 4°C.

The temperature of the water is 4 °C.

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