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1 year ago

Which particle model diagram represents xenon at stp?

1 answer:

3 0

The model that should show the corresct representation of xenon gas is one in which the gas molecules are isolated and monoatomic.

<h3>What is a noble gas?</h3>

A noble gas is a member of group 18 of the periodic table. Noble gases are known not to interact with each other and occur as monoatomic particles.

The images are not shown here hence the question is incomplete. However, we do know that any of the models that show individual monoatomic particles is a representation of xenon gas.

Learn more about noble gas: brainly.com/question/2094768

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Information gathered by a scientist abut the toxicity of chemical X and chemical Y showed that they had individual safe limits f

cupoosta [38]

Answer:

Synergism

Explanation:

Synergism -

It refers to the process of interaction between the any two or more chemicals , i.e. , combining the chemicals increases the effect of the chemical , is referred to as synergism .

The individual drug or chemical might not be strong or harmful in any respect , but combining it with various other drugs or chemical , might make it very strong or harmful .

Hence ,from the given scenario of the question ,

The correct answer is synergism .

3 0

2 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Lera25 [3.4K]

Answer:

NH3

Explanation:

2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)

So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.

n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2

4 0

2 years ago

A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3p

timurjin [86]

Answer:

55.75g

Explanation:

From

m/M = CV

Where

m= required mass of solute

M= molar mass of solute

C= concentration of solution

V= volume of solution=675ml

Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1

Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles

Since 1 mole of Na3PO4 contains 3 moles of Na+

It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles

mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g

3 0

3 years ago

please help me with Chem I ONLY HAVE 5 MINUTES if methane gas (CH4) flows at a rate of 0.25L/s, how many grams of methane gas wi

Nookie1986 [14]

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

<em>Volume Methane:</em>

3600s * (0.25L / s) = 900L Methane

<em>Moles methane:</em>

PV = nRT; PV / RT = n

<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

<em>Mass methane:</em>

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>

5 0

2 years ago

The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3

jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 $in^{3}$

Given density = 0.242 g/ $cm^{3}$

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is $in^{3}$ and unit of volume in density is $cm^{3}$ , so first we need to change the unit of volume from $in^{3}$ to $cm^{3}$ so that the volume units will cancel out, leaving only the mass units.