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3 years ago

What particles do during phase changes

Chemistry

1 answer:

andrey2020 [161]3 years ago

8 0

Explanation:

Hence in this experiment, the importance of experiment is being tested, as it is added only in one set of plant.

A manipulated variable is an independent variable whose amount can be changed during an experiment.

Here the fertilizer is a manipulated variable.

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Which type of bond is not backed by any specific collateral, such as land or equipment?

GrogVix [38]

Answer:

Explanation:

looking at both equity

and mortgage bond fixtures

I would rather say secured bond fit into not being backed by any specific collateral

7 0

1 year ago

A dental assistant is preparing to use a chemical that has a number 2 in a blue triangle on the chemical label. Before using the

Effectus [21]

Labels must clearly identify the chemical and include information on its hazards, plus instructions and information on its safe storage, handling, use, and disposal.

What does the information on the chemical label indicate?

The chemical label indicates the following information.

Product Identifier: The method used to identify the chemical (in the ingredient disclosure). On the label, include the complete chemical name or number that is used for the hazardous product. Both the label and SDS must have the same information. Abbreviations for chemicals are not accepted.

Signal Words: Assigned to a GHS hazard class and category, "Danger" or "Warning" are used to underline hazards and show the relative level of severity of the hazard.

Danger Statements: Typical expressions that are associated with a hazard class and category and that define the hazard's characteristics. On a label, there must be mention of every danger that applies.

Precautionary Statements: A term that describes actions that should be performed to reduce or eliminate negative effects brought on by exposure to or inappropriate storage of a substance.

Symbols (hazard pictograms): Convey information about environmental, physical, and health hazards that are categorized and allocated to a GHS hazard class.

Learn more about the chemical label here:

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5 0

2 years ago

You use a _______ when you are changing compounds

zvonat [6]

Answer:

snetence

Explanation:

nikkak youk dxmbk aslk istgk

3 0

2 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago