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2 years ago

How many grams of nitrogen are in a sample of Al(NO3)3 that contains 364.0 g of aluminum? Answer: ________ grams of nitrogen.

Chemistry

1 answer:

saw5 [17]2 years ago

7 0

<u>Answer:</u> The mass of nitrogen that is present for given amount of aluminium is 566.22 g.

<u>Explanation: </u>

We are given:

Mass of aluminium = 364 grams

The chemical formula of aluminium nitrate is $Al(NO_3)_3$

Molar mass of nitrogen = 14 g/mol

Molar mass of aluminium = 27 g/mol

In 1 mole of aluminium nitrate, 27 grams of aluminium is combining with 42 grams of nitrogen.

So, 364 grams of aluminium will combine with = $\frac{42}{27}\times 364=566.22g$ of nitrogen.

Hence, the mass of nitrogen that is present for given amount of aluminium is 566.22 g.

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A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

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2 years ago

A student needed exactly 45.3ml of a solution. what piece of glassware should that student use? justify your choice.

Darina [25.2K]

The student should use the graduated cylinder. A graduated is the most common laboratory glassware when measuring volumes. It has calibrations by 1, 0.5 or 0.1 depending on the maximum volume. You have to make sure though, that you measure the volume by looking at the lower meniscus of the liquid at eye level.

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1 year ago

In some chemical reactions, electrons are transferred from one substance to another, but there is no net gain or loss of electro

Natali [406]

Answer and Explanation:

The equation that depicts oxidation of neutral atom A is shown below:

$\rightarrow A^+ + e^-$

This is because one species is losing electrons due to oxidation. The species possesses positively charged after losing electrons, the magnitude of which is proportional to the number of electrons lost.

The net charge will be equivalent on both sides of the equation, too.

Therefore all other options are not correct

The equation that depicts the decline of neutral atom X is

$X + e^- \rightarrow X^-$

It is how a cell gains electrons by reduction. The species obtains a negative charge upon possessing electrons, whose magnitude is equivalent to the amount of electrons gained.

The net charge will be equivalent on both sides of the equation, too.

Therefore all other options are not correct

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2 years ago

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the a

UkoKoshka [18]

Answer:

$m_{Ga}=0.550gGa$

Explanation:

Hello!

In this case, since the applied current for the 50.0 mins provides the following charge to the system:

$q=0.760\frac{C}{s}*50.0min*\frac{60s}{1min}=2,280C$

As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:

$n_{e^-}=2,280C*\frac{1mole^-}{96,485C}=0.0236mole^-$

Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:

$m_{Ga}=0.0236mole^-*\frac{1molGa}{3mole^-}*\frac{69.72gGa}{1molGa} \\\\m_{Ga}=0.550gGa$

Best regards!

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1 year ago