A gas sample of 5 moles, has a volume of 95 L. How many moles of the same gas should I add to obtain a volume of 133 L at the same temperature and pressure.

Answer:

There are no options provided dude,

But i guess the  answer will be a metal with valency 2 for sure as the subscript given for N in 'M3N2' is 2 so the valency of the metal u need to select will be 2 for sure

It can be magnesium or some other if the provided options in real question has Mg then its the answer

Answer:31 days

Explanation:

3500kcal/lb x 5lb =17500

1800-1230=570

17500/570=30.70

Round up =31

This question involves the concepts of energy, weight, and time.

It will take Charles "9.7 days" to lose 5.0 lb.

Since the loss of 1.0 lb weight requires expending 3500 kcal energy. Therefore, by using the unitary method,

1.0 lb loss = 3500 kcal

(5)(1.0) lb loss = (5)(3500 kcal)

5.0 lb loss = 17500 kcal ----- eqn(1)

Now, the time required to consume 1800 kcal is 1 day. Hence, by the unitary method,

1800 kcal = 1 day

1 kcal = [tex]\frac{1}{1800}\ day[/tex]

17500 kcal = [tex]\frac{17500}{1800}[/tex] day

17500 kcal = 9.7 days ------ eqn(2)

Comparing eqn (1) and eqn (2)

5.0 lb loss = 9.7 days

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Answer:

Cu2 + 4 NH3 = Cu (NH3) 42+

so - - >

[Cu (NH3) 6] ^ 2+

Explanation:

Answer:

The element is a nonmetal

Explanation:

Elements are broadly classified into metals and non metals. Metals conduct electricity while non metals do not conduct electricity.

If we look at this scenario described in the question, we can easily decipher that the element is a nonmetal because the LED stays dark. The LED should have been lit if electricity was passed through the element in question.

Hence, the element is a non metal.

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) →  SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

The mass allowance of Vitamin C for children aged 4-8 years is equal to 0.025 grams.

A mole can be defined as a standard unit that can be utilized to evaluate the number of entities such as atoms, molecules, ions, or other particular particles in a particular amount of the substance.

The number of elementary entities present in one mole of any chemical substance was found to be equal to 6.023 × 10²³ which is also known as the Avogadro number.

Given, the number of moles of vitamin C = 0.000142 moles

Given, the molecular formula of Vitamin C is C₆H₈O₆.

The mass of one mole of C₆H₈O₆ = 176 g

One mole of Vitamin C has mass = 176 g

0.000142 mol of Vitamin C has mass  = 0.000142×176 = 0.025 g

Therefore, the mass of 0.000142 mol of Vitamin C is 0.025 g.

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Answer:

2.68 cm^3

Explanation:

Density= Mass/Volume

so...

8.96 g/cm^3 = 24.01 g/ V

and then u solve so it would be 2.68 cm ^3

((:

Answer:

Correct answer is D : T <  1980°C

Explanation:

Correct answer is D:T< 1980°C

Perovskite proton conductors belong to the class of high temperature proton conductors (HTPCs) , solid that conduct electricity by transporting H⁺ ions (protons ) at temperatures above ambient , typically 400 - 1000°C

So, we get T < 1980°C

Answer:

nice handwrtting

Explanation:

Answer:

[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2

Explanation:

2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression in terms of concentrations is:

Kc=[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2.

The equilibrium expression for the given reaction can be written in terms of equilibrium constant which is the ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

Equilibrium is a state for a reversible reaction where, the rate of forward reaction is equal to the rate of backward reaction. The rate of a reaction is the rate of decrease in the concentration of reactants or the rate of increase in the concentration of the products.

The given reaction at equilibrium state is written as:

[tex]\rm 2H_{2}S (g)\leftrightharpoons 2H_{2} (g)+ S_{2}(g)[/tex]

The equilibrium constant Kb is ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

[tex]Kb = \rm \frac{[H_{2}S]^{2}}{[H_{2}]^{2} [S_{2}]}[/tex]

The rate of the reaction will be r = Kb [H₂]² [S₂].

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Answer:

Empirical formula

Explanation:

The empirical formula of a compound is the simplest whole number ratio of atoms of each element in the compound. It is determined using data from experiments and therefore empirical.

Answer:

wow!

5. C

6. B

7. B

8. A

Explanation:

Answer:

The conditions are

1) Small enough

2) Electronegative atom

3) highly electronegative

4) lone pair of electrons

The correct statement therefore is

It needs to be small enough not to bump into other atoms when approaching the 1s orbital of the hydrogen, it needs to carry at least one electronegative atom, it needs to be highly electronegative enough to create a delta on the connected hydrogen, and it needs to have at least one lone pair of electrons.

Explanation:

Hydrogen bonding is a type of intermolecular bond that occurs between the partial positive charge (delta) on a hydrogen atom bonded to a small highly electronegative element (like nitrogen, oxygen or fluorine) and the free electrons on another electronegative element of another molecule.

The hydrogen atom with the partial positive charge (delta) is known as the hydrogen bond donor, while the electronegative element, carrying lone electrons is called the hydrogen bond acceptor.

Let's take a deeper look at these terms:

1) Hydrogen bond donor

Using water (H₂O) as an example, the high electronegativity of the oxygen atom covalently bonded to the hydrogen atom draws the lone electron in the 1s orbital of the hydrogen atom, creating a partial positive charge (d⁺) on the hydrogen atom. This is what happens within one water molecule

2) Hydrogen bond acceptor

When two or more molecules of water interact, the partial positive charge (d⁺) on the hydrogen atom of one molecule, is attracted to the valence or free electrons on the oxygen atom of a nearby molecule of water thus creating a dipole-dipole intermolecular bond known as a hydrogen bond.

For the hydrogen bond to be effective, the electronegative atom bonded to the hydrogen acting as the hydrogen bond donor in the first water molecule needs to be small enough so as not to disrupt the 1s orbital of the hydrogen atom. The smaller the size of the electronegative atom, the stronger the partial negative charge created on the hydrogen atom.

The valence or free pair of electrons on the electronegative (oxygen) atom of the second molecule of water (hydrogen bond acceptor) is what attracts the partial positive charge on the hydrogen atom to create the hydrogen bond

Answer:

C8H16O4

Explanation:

C2H4O= 24+4+16

44

n=molar mass/empirical formula

n=176.21/44

=4

Therefore

Molar Formula= (C2H4O)4=C8H16O4

Answer: The molar mass of menthol is 156.15 g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=2.84^0C[/tex] = Depression in freezing point

[tex]K_f[/tex] = freezing point constant = [tex]20.8^0C/m[/tex]

m= molality

[tex]\Delta T_f=K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (cyclohexane)= 25.0 g = 0.025 kg  

Molar mass of solute (menthol) = ?

Mass of solute (menthol) = 0.533 g

[tex]2.84^0C=20.8\times \frac{0.533}{xg/mol\times 0.025}[/tex]

[tex]x=156.15g/mol[/tex]

The molar mass of menthol is 156.15 g/mol

Answer:

% = 33.83%

Explanation:

To do a better understanding of this, we can treat the mixture of the combustion as two separate reactions, in that way, we can have an idea of what is happening and how to calculate the mass percent.

So the combustion reactions in this mixture are:

2C₂H₆ + 7O₂ ---------> 4CO₂ + 6H₂O

C₃H₈ + 5O₂ ----------> 3CO₂ + 4H₂O

Now that we have both reactions (And balanced) we can hace an idea of the mole ratio between every compound in the mix.

For practical purposes, let's call "a" the mass of ethane, and "b" the mass of propane. The innitial mix have 9.725 g, so this means that:

a + b = 9.725 g    (1)

Now that we have this, we can write a relation between the moles of oxygen and the moles of the gases. If we have 1.115 moles of oxygen, and also know the mole ratio of oxygen to "a" and "b", so:

moles O₂ = moles a (moles O₂/moles a) + moles b (moles O₂/moles b)   (2)

And we know that moles a and moles b are:

moles a = a / MW

moles b = b / MW

The MW of a is 30 g/mol and the MW of b is 44 g/mol

Replacing the given data we have:

1.115 moles O₂ = (a/30)(7 moles O₂/2 moles a) + (b/44)(5 moles O₂/1 mole b)

1.115 moles O₂ = (0.1167a) moles O₂ + (0.1136b) moles O₂

To keep solving this, we can use expression (1) to solve for b, and then, replace here and have only one equation with 1 incognite:

a + b = 9.725 g  

b = 9.725 - a   (3)

Replacing above we have:

1.115 = 0.1167a + 0.1136(9.725 - a)

1.115 = 0.1167a + (1.1048 - 0.1136a)

1.115 - 1.1048 = 0.1167a - 0.1136a

0.0102 = 0.0031a

a = 3.29 g

Now, that we have the mass of the ethane, we can calculate the mass percent:

% = (3.29 / 9.725) * 100

Hope this helps

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

Answer:

[tex]atoms= 1.204x10^{24}atoms[/tex]

Explanation:

Hello!

In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:

[tex]atoms=2.00mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms= 1.204x10^{24}atoms[/tex]

Best regards!

Answer:

2

Explanation

It seems logical

Answer:

Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction  N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction  N₂ : He : CH₄ : C₂H₆ : : .5357  : 1.25 : 3.75 : .67

Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057

mole fraction of N₂ =  .5357 / 6.2057 = .0877

mole fraction of He = 1.25 / 6.2057 = .20

mole fraction of CH₄ = 3.75 / 6.2057 = .6043

mole fraction of C₂H₆ = .67 / 6.2057 = .108

Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa  x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa  x .20  = 240 kPa

Partial pressure of CH₄ = 1200 kPa  x  .6043  = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa  x .108    = 129.6 kPa

Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.

Explanation:

The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.

Answer:

b. temperature difference

Answer:

increase the chemical rate

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

Answer:

d. Mass is the measure of the earth’s gravitational attraction of a body

This is FALSE. Mass is the measure of matter than an object contains.

Explanation:

b. An atom is the smallest particle of an element that still contains properties of the original element.

This is True

c. Gases can be colorless or colored

This is True

d. Mass is the measure of the earth’s gravitational attraction of a body

This is FALSE. Mass is the measure of matter than an object contains.

e. A pure substance has a fixed composition.

This is True.

Answer:

0.73g/cm^3

Explanation:

d=m/v

d=11/15

d=0.73

Answer: Octane will be used completely.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of octane}=\frac{2.3g}{114g/mol}=0.0202moles[/tex]

[tex]\text{Moles of oxygen}=\frac{12.4g}{32g/mol}=0.388moles[/tex]

The balanced chemical reaction will be

[tex]2C_8H_{18}+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]  

According to stoichiometry :

2 moles of octane require = 25 moles of [tex]O_2[/tex]

Thus 0.0202 moles of octane will require=[tex]\frac{25}{2}\times 0.0202=0.2525moles[/tex]  of [tex]O_2[/tex]

Thus octane is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.  

Thus octane will be used completely.