Calculate the mass of a liquid with a density of 3.2 g/ml and a volume of 25 ml

. A sample of crude oil has a density of 0.87 g/mL. What volume (in liters) does a 3.6 kg sample of this oil occupy

gayaneshka [121]

Answer:

The volume is 4.13793 L

Explanation:

Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:

$density=\frac{mass}{volume}$

Density is a characteristic property of every body or substance.

The most commonly used units of density are $\frac{kg}{m^{3} }$ or $\frac{g}{cm^{3} }$ for solids, and $\frac{kg}{L}$ or $\frac{g}{mL}$ for liquids and gases.

In this case, you know:

density= 0.87 $\frac{g}{mL}$
mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)
volume= ?

Replacing:

$0.87\frac{g}{mL} =\frac{3,600 g}{volume}$

Solving:

$volume =\frac{3,600 g}{0.87\frac{g}{mL}}$

volume= 4,137.93 mL

Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L

<u><em>The volume is 4.13793 L</em></u>

5 0

3 years ago

explain why the hydrogen atoms in a hydrogen gas molecule form nonpolar covalent bonds but oxygen and hydrogen atoms in water mo

marishachu [46]

Answer:

Explanation:

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.

The atom with larger electronegativity attract the electron pair more towards itself and becomes partial negative while the other atom becomes partial positive.

Non polar covalent bond:

It is the bond where both bonded atoms share the pair of electron equally.

For example:

Hydrogen gas (H₂) is non polar covalent compound because the electronegativity of both bonded atoms are same. No poles are created that's why this is non polar covalent compound.

Polar covalent bond:

It is the bond where both bonded atoms share the pair of electron unequally.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive and bond is polar.

7 0

3 years ago

Dana is trying to determine if an unknown sample is a pure substance or not. The sample has a cloudy white color, and it seems e

Firdavs [7]

Answer:

Dana filtered the sample and larger granules of the sample were left behind.

Explanation:

If a substance is pure, it will have a uniform composition throughout. It will not separate into particles of various sizes.

One of the characteristics of pure substances is that they are homogeneous. A mixture is definitely made up of particles of various sizes.

Since the particles was filtered and larger granules were left behind, the sample has been separated by a physical method (filtration). Only a mixture can be separated by physical methods. It is not a pure substance.

5 0

2 years ago

When 33.3 grams of propane (C3H8) undergoes combustion, what is the theoretical yield of water in grams? The molar mass of p

aev [14]

C3H8+ 5 O2 --> 3 CO2 + 4 H2O
44 g. --------> 72 g
33.3 g. --------> x
$x = \frac{33.3 \times 72}{44} \\ x = 54.5 \: g$
Answer: The theoretical yield of H2O is 54.5

3 0

3 years ago

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Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

2 years ago

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