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3 years ago

What is the density of nitrogen gas if 0.186 g occupies a volume of 110 mL?

Chemistry

1 answer:

Anvisha [2.4K]3 years ago

4 0

Answer:

$d=1.69\times 10^{-3}\ g/mL$

Explanation:

Given that,

Mass, m = 0.186 g

Volume occupies, V = 110 mL

We need to find the density of Nitrogen gas. The mass per unit volume is called density of an object. It is given by the formula as follows :

$d=\dfrac{m}{V}\\\\d=\dfrac{0.186\ g}{110\ mL}\\\\=1.69\times 10^{-3}\ g/mL$

So, the required density is $1.69\times 10^{-3}\ g/mL$ .

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Explanation:

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3 years ago

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Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample

goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=2:2=1:1$

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

$n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol$

Similarly, number of moles of oxygen will be:

$n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol$

The ratio of number of moles of carbon and oxygen will be:

$C:O=n_{C}:n_{O}=3:3=1:1$

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.

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3 years ago

The density of aluminum is 2.70 g/cm'. If the volume of a piece of aluminum is 90.4 cm', what is

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Answer:

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Explanation:

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3 years ago

Standard reduction potentials for nickel(II) and copper(II) The standard reduction potential for a substance indicates how readi

AleksandrR [38]

Answer:

The standard potential, E cell, for this galvanic cell is 0.5670V

Explanation:

Ni²⁺(aq) + 2e⁻ → Ni(s) E red = - 0.23V ANODE

Cu²⁺(aq) + 2e- → Cu(s) E red = + 0.337V CATHODE

ΔE° = E cathode - E anode

ΔE° = 0.337V - (0.23V) = 0.5670 V

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3 years ago

Enter your answer in the provided box. From the data below, calculate the total heat (in J) needed to convert 0.304 mol of gaseo

diamong [38]

Answer:

-35,281.5 J

Explanation:

To convert the gaseous ethanol to liquid ethanol, three steps will occur. First, it will lose heat and the temperature will decrease until its boiling point, so from 300.0°C to 78.5°C. Thus, more heat will be lost, but now, with the temperature constant, so the gas will be converted to liquid. And then, the liquid will lose heat to decrease the temperature from 78.5°C to 25.0°C.

The total heat loss is the sum of the heats of each step. Because the heat is being removed from the system, it's negative. The first and last step occurs with a change in temperature, and so the heat is calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat of the gas (first step) or liquid (last step), and ΔT the temperature variation (final - initial). The mass of ethanol is the molar mass 46.07 g/mol multiplied by the number of moles, so:

m = 46.07 * 0.304 = 14.00 g

The second step occurs without a change in temperature, and the heat is then:

Q = -n*ΔH°vap

Where n is the number of moles, ΔH°vap is the heat of vaporization, and the minus signal indicates that the heat is being lost. Then, the heat of each step is:

Q1 = 14.00*1.43*(78.5 - 300,0) = -4434.43 J

Q2 = -0.304*40.5 = -12.312 kJ = -12312 J

Q3 = 14.00*2.45*(25.0 - 78.5) = -1835.05 J

Q = Q1 + Q2 + Q3

Q = -35,281.5 J

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3 years ago