what is the mass in grams of one mole of He?

Answer:

Full moon.

Explanation:

This diagram shows a full moon because the sunlight is shining directly onto the side of the moon facing the earth in this instance.

Answer:

It is A

Explanation:

Answer:

THE NEW VOLUME OF THE GAS SAMPLE IS 40 L AT 1 atm AND 400 K

Explanation:

Using general gas equation which combines Boyle's and Charles' law, the volume of the gas sample at the new pressure and temperature can be obtained.

P1V1/T1 = P2V2/T2

P1 = Initial pressure = 2 atm

V1 = Initial volume = 15 L

T1 = Initial temperature = 300 K

P2 = Final pressure = 1 atm

T2 = Final temperature = 400 K

V2 = Final volume = unknown

Substituting the values into the equation, we have;

V2 = P1V1T2 / P2 T1

V2 = 2 * 15 * 400 / 1 * 300

V2 = 12 000/ 300

V2 = 40 L

The new volume after the pressure was reduced to 1 atm and the temperature increased to 400 K is 40 L.

Answer: Radioactive waste

Explanation:

The nuclear fission reaction consists of heavy atomic particles or heavy nucleus, like plutonium and uranium and in radioactive heavy metals. In the fission reaction the nucleus get split into equal masses of particles. This process is associated with release of large amount of energy. The fission of radioactive waste can cause deadly mutations in living beings.      

Answer:

B. 47.3

Explanation:

How compact a certain object or person is

Answer:

Density is a measure of mass per unit of volume.The average density of an object equal its total mass divided by its total volume.

Answer:

Type of reactions :

Answer:

I has 5 valance electrons it wants 10 and 3 bonds

Explanation:

Answer:

[tex]NaSO_{4}[/tex]

Explanation:

Asúmase que existen 100 gramos de compuesto, las cantidades molares de cada componente son, respectivamente: (Let assume that exist 100 grams of the composite, the molar quantities of each component are, respectively:)

Na

[tex]n = \frac{19.3\,g}{22.99\,\frac{g}{mole} }[/tex]

[tex]n = 0.839\,moles[/tex]

S

[tex]n = \frac{26.9\,g}{32.065\,\frac{g}{mole} }[/tex]

[tex]n = 0.839\,moles[/tex]

O

[tex]n = \frac{53.8\,g}{15.999\,\frac{g}{mole} }[/tex]

[tex]n = 3.363\,moles[/tex]

La fórmula molecular empírica es (The empirical molecular formula is):

[tex](Na_{0.839}S_{0.839}O_{3.363})_{1.192}[/tex]

[tex]NaSO_{4}[/tex]

Answer: 1. [tex]NH_3[/tex] increases.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+energy[/tex]

If the concentration of [tex]N_2[/tex] is increased , according to the Le-Chatlier's principle, the reaction has to shift to right or forward direction. In order to do that  the concentration of products has to increase.

Thus the concentration of [tex]NH_3[/tex] increases.

Answer:

601mmHg.

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 760mmHg

Initial temperature (T1) = 100°C = 100°C + 273 = 373K

Final temperature (T2) = 22°C = 22°C + 273 = 295K

Final pressure (P2) =..?

Step 2:

Determination of the new pressure at which the water will boil.

The new pressure on the at which the water will boil can be obtained as follow:

P1/T1 = P2/T2

760/373 = P2/295

Cross multiply to express in linear form

373 x P2 = 760 x 295

Divide both side by 373

P2 = (760 x 295) / 373

P2 = 601mmHg

Therefore, the new pressure at which the water will boil is approximately 601mmHg.

Answer:

C. 18-20 mm

Explanation:

Answer: 305 g of [tex]Be_3(PO_4)_2[/tex] will be produced from 38 grams  of beryllium oxide

Explanation:

To calculate the moles :

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

[tex]\text{Moles of} BeO=\frac{38g}{25g/mol}=1.52moles[/tex]

[tex]3BeO+2FePO_4\rightarrow Be_3(PO_4)_2+Fe_2O_3[/tex]

According to stoichiometry :

3 moles of [tex]BeO[/tex] produce = 1 mole of [tex]Be_3(PO_4)_2[/tex]

Thus 1.52 moles of [tex]BeO[/tex] will produce =[tex]\frac{1}{3}\times 1.52=0.507moles[/tex]  of [tex]Be_3(PO_4)_2[/tex]

Mass of [tex]Be_3(PO_4)_2=moles\times {\text {Molar mass}}=0.507moles\times 602g/mol=305g[/tex]

Thus 305 g of [tex]Be_3(PO_4)_2[/tex] will be produced from 38 grams  of beryllium oxide

1107 Joule is the amount of energy which is needed to change 88.0 g of solid benzene at 5.53°C into liquid benzene, also at 5.53°C.

In order to perform work and to produce heat and light, energy must be delivered to a body or to an external physical system. Energy is a quantitative property. Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be created or destroyed. 

A moving object's kinetic energy, an object's potential energy, an object's elastic energy, chemical energy linked to chemical reactions, electromagnetic radiation's radiant energy, and the internal energy of a thermodynamic system are examples of common kinds of energy.

mole of benzene = 85.2/78.11 =1.09mol

1 mole of benzene requires  10.19 kJ/mol energy

1.09 mole of benzene requires 1.09× 10.19 kJ/mol = 11.107kJ/mol energy

                                                                                  = 1107 Joule energy

Therefore,  1107 Joule is the amount of energy which is needed to change 88.0 g of solid benzene at 5.53°C into liquid benzene, also at 5.53°C.

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Answer:

% yield = 44.5 %

Explanation:

Step 1: Data given

Mass of Mg3N2 = 100.0 grams

Mass of H2O = 75.0 grams

Mass of NH3 = 15.0 grams

Molar mass of Mg3N2 = 100.95 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

Mg3N2 + 3H2O → 3MgO + 2NH3

Step 3: Calculate moles

Moles = mass / molar mass

Moles Mg3N2 = 100 grams / 100.95 g/mol

Moles Mg3N2 = 0.99 moles

Moles H2O = 75.0 grams / 18.02 g/mol

Moles H2O = 4.16 moles

Moles NH3 = 15.0 grams / 17.03 g/mol

Moles NH3 = 0.88 moles

Step 4: Calculate the limiting reactant

For 1 mol Mg3N2 we need 3 moles H2O to produce 3 moles MgO and 2 moles NH3

Mg3N2 is the limiting reactant. It will completely be consumed (0.99 moles)

H2O is in excess. There will react 3*0.99 = 2.97 moles

There will be produced 2*0.99 = 1.98 moles of NH3

Step 5: calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 1.98 moles * 17.03 g/mol

Mass of NH3 = 33.7 grams

Step 6: Calculate the % yield

%yield = (actual yield/ theoretical yield) * 100 %

% yield = (15.0 grams / 33.7) * 100 %

% yield = 44.5 %

The percentage yield of ammonia in the given reaction has been 44.44%.

The balanced chemical reaction can be:

[tex]\rm Mg_3N_2\;+\;3\;H_2O\;\rightarrow\;3\;MgO\;+\;2\;NH_3[/tex]

The moles can be given as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of [tex]\rm Mg_3N_2[/tex] = 0.99 mol

Moles of [tex]\rm H_2O[/tex] = 4.16 mol

Moles of [tex]\rm NH_3[/tex] = 0.88 mol

The limiting reactant can be:

From the balanced equation for 3 moles of water, there has been 1 mole of magnesium nitrite. So, for 4.16 moles of water, there will be 1.38 moles of [tex]\rm Mg_3N_2[/tex]. Since the amount of [tex]\rm Mg_3N_2[/tex] has been completely consumed in the reaction, [tex]\rm Mg_3N_2[/tex] has been the limiting reactant.

So, the moles of ammonia produced =

1 moles [tex]\rm Mg_3N_2[/tex] = 2 moles ammonia.

0.99 moles [tex]\rm Mg_3N_2[/tex] = 0.99 [tex]\times[/tex] 2 moles [tex]\rm NH_3[/tex]

0.99 moles [tex]\rm Mg_3N_2[/tex] = 1.98 moles of ammonia.

The theoretical yield of ammonia = 1.98 moles

The actual yield of ammonia = 0.88 mol.

% Yield of Ammonia = [tex]\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100[/tex]

% Yield of Ammonia = [tex]\rm \dfrac{0.88}{1.98}\;\times\;100[/tex]

% Yield of Ammonia = 44.44 %

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Answer:  3597 kJ of heat

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 5.00 atm

V = Volume of gas = 8.00 L

n = number of moles = ?

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]25.0^0C=(25.0+273)K=298K[/tex]

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{5.00atm\times 8.00L}{0.0821 L atm/K mol\times 298K}=1.63moles[/tex]

As it is given :

1 mole of propane produces = 2220 kJ of heat

Thus 1.63 moles of propane produces = [tex]\frac{2200}{1}\times 1.63=3597kJ[/tex]

Thus 3597 kJ of heat is produced

Answer:

No they are not

Explanation:

Energy levels of electrons have fixed distance from the nucleus of an atom where electrons may be found.

orbitals is the path that electrons revolve

Answer:

Option a. = 0.01 M

Explanation:

To do this, we need to gather the data:

E = 2.21 V

[Cl⁻] = 0.1 M

And the Redox reaction taking place is the following:

Zn(s) + Cl₂(g) <-------> Zn²⁺(aq) + 2Cl⁻(aq)       Q = [Zn] [Cl]²

E° Cl⁻/Cl₂ = 1.36 V

E° Zn/Zn²⁺ = -0.76 V

According to this, the expression to use will be the Nernst equation, and we can assume we are working at 25 °C, therefore, the Nernst equation will be:

E = E° - (0.059/n) logQ

E = E° - (0.059/n) ln([Cl⁻]² * [Zn²⁺])   (1)

From there, we can solve for Zn later.

First, we need to write the semi equation of oxidation and reduction, and get the standard potential of the cell:

Zn(s) --------> Zn²⁺(aq) + 2e⁻       E₁° = 0.76 V

Cl₂(g) + 2e⁻ -----------> 2Cl⁻(aq)   E₂° = 1.36 V

---------------------------------------------------------------

Zn(s) + Cl₂(g) -------> Zn²⁺(aq) + 2Cl⁻(aq)    E° = 0.76 + 1.36 = 2.12 V

Now, let's replace in (1) and then, solve for [Zn]:

2.21 = 2.12 - (0.059/2) log ([0.1]² * [Zn])

2.21 - 2.12 = -0.0295 log (0.01[Zn])

- 0.09 / 0.0295 = log (0.01[Zn])

-3.0508 = log (0.01[Zn])

10^(-3.0508) = 0.01[Zn]

8.8961x10⁻⁴ = 0.01[Zn]

[Zn²⁺] = 0.08896 M

This value can be rounded to 0.1 M. so the correct option will be option A.

Answer:

false

Explanation:

Answer:

evaporation

Explanation:

Answer:

Boiling

Explanation:

When a substance transitions from liquid to gas, it's called evaporation if it's on the surface, and boiling if it's all throughout the liquid.

Picture boiling water in a pot: it starts on the bottom, and the bubbles go toward the top. Evaporation starts on the surface.

Answer:

B

Explanation:

I’m not really great at chemistry but I truly tried my best

Answer:

i would say (c)

Explanation:

Answer:c

Explanation:

Answer: Yes, the student is right, one should use pellets of the reactant should be heated and stirred for mixing properly.

Explanation:

In case of smaller particles the surface area that is being exposed increases and the due to this the reaction occurs faster.

Increasing the temperature of the temperature, increases the kinetic energy of the particles which helps in easy mixing of the particles.

The collision in between the particles also increase while stirring and thus the rate of reaction increases.

So, the heating and stirring is more preferred over powered reactant for making a solution quickly.

Answer:

Silver (Ag)

Explanation:

The electronic configuration of copper is shown below;

[Kr]4d10 5s1

We can see that there is only one 5s1 electron. Hence Ag^+ tends to display a pseudo noble gas configuration. This pseudo noble gas configuration explains why silver is prevalent in the +1 oxidation state.

The other transition metals have many stable oxidation states found in nature. Chromium is observed both in +3 and +6 oxidation states. Iron is found in +2 and +3 oxidation states and copper is mostly stable in the +2 oxidation state since the +1 oxidation state readily disproportionate.

Hence silver tends to have only a single ionic charge for reasons aptly stated above.

Answer:

water and carbon dioxide

Explanation:

Chemical weathering is the process of changing the composition of a mineral in order to break it down. The two agents that are the biggest proponents of chemical weathering are water and carbon dioxide. Water is used in a process called Hydrolysis which uses it to break down substances such feldspar in granite rocks changing to clay. While Carbon Dioxide in the atmosphere dissolves in rainwater forming carbonic acid, which slowly dissolve rocks when it comes into contact with them.

Answer:

62,790J

Explanation:

Q=mcT

= 250g*4.186J/gc*80-20

=250*4.186*60=62,790J

Answer:

Explanation:

SO; If we assume that:

P should be the diffusion of oxygen towards the surface ; &

Q should be the  diffusion of carbondioxide away from the surface.

Then  the total molar flux of oxygen is illustrated by :

[tex]Na,x = - cD_{PQ}\frac{dy_P}{dr} +y_P(NP,x + N_Q,x)[/tex]

where;

r is the radial distance from the center of the carbon particle.

Since ;

[tex]N_P,x = - N_Q, x[/tex] ; we have:

[tex]Na,x = - cD_{PQ}\frac{dy_P}{dr}[/tex]

The system is not steady state and the molar flux is not independent of r because the area of mass transfer [tex]4\pi r^{2}[/tex] is not a constant term.

Therefore,  using quasi steady state assumption, the mass transfer rate   [tex]4\pi r^{2}N_{P,x}[/tex] is assumed to be independent of r at any instant of time.

[tex]W_{P}=4\pi r^{2}N_{P,x}[/tex]

[tex]W_{P}=-4\pi r^{2}cD_{PQ}\frac{dy_{P}}{dr}[/tex]        

       = constant

The oxygen concentration at the surface of the coal particle [tex]yP,R[/tex] will be calculated from the reaction at the surface.

The mole fraction of oxygen at a location far from pellet is 1.

Thus, separating the variables and integrating result into  the following:

[tex]W_{P}\int_{R}^{\infty} \frac{dr}{r^{2}}=-4\pi[/tex]

[tex]r^{2}cD_{PQ}\int_{y_{P,R}}^{y_{P,\infty }}dy_{P}[/tex]

[tex]-W_{P}\frac{1}{r}\mid ^{\infty }_{R}= -4\pi cD_{PQ}(y_{P,\infty }-y_{P,R})[/tex]

[tex]=> W_{P}= - 4\pi cD_{PQ}(1-y_{P,R})R[/tex]

The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed by the chemical reaction.

[tex]W_{P} = 4 \pi R^2R"[/tex]

[tex]W_{P}= 4\pi R^{2}k_{1}"C_{O_{2}}\mid _{R}[/tex]

[tex]W_{P}= 4\pi R^{2}k_{1}"c y _{P,R}[/tex]

[tex]-4\pi cD_{PQ}(1-y_{P,R})R= - 4\pi R^{2}k_{1}"c y _{P,R}[/tex]

[tex]y_{P,R}=\frac{D_{PQ}}{D_{PQ}+Rk_{1}}[/tex]

[tex]y_{P,R}=\frac{1.7 \times 10^{-4}}{1.7\times 10^{-4}+10^{-3}\times 0.1}[/tex]

[tex]\mathbf{= 0.631}[/tex]

Obtaining the total gas concentration from the ideal gas law; we have the following:

where;

R= [tex]0.082m^3atm/kmolK[/tex]

[tex]c=\frac{P}{RT} \\ \\ c=\frac{1}{0.082\times 1450} \\ \\ = 0.008405kmol/m^3[/tex]

The steady state [tex]O_2[/tex] molar consumption rate is:

[tex]W_{P}= -4\pi cD_{PQ}(1-y_{P,R})R[/tex]

[tex]W_{P}= -4\pi (0.008405)(1.7\times 10^{-4})(1-0.631)(10^{-3})[/tex]

[tex]W_{P}= - 6.66\times 10^{-9}kmol/s[/tex]

Answer:

[tex]3 Mg(s) + N_2(g) = Mg_3N_2(s)[/tex]

Explanation:

The valance of magnesium atom is  [tex]+2[/tex]

While the valence of nitrogen atom is [tex]- 3[/tex]

Let us first write the first  half-reactions

[tex]Mg ---> Mg^{2+} + 2e^{-}[/tex]

The second half reaction is

[tex]2N + 6e^- -----> N_2[/tex]

Adding the above two reactions and writing the final reaction, we get -

[tex]Mg + N2 = Mg_3N_2[/tex]

The balance equation is

[tex]3 Mg(s) + N_2(g) = Mg_3N_2(s)[/tex]

Answer:7

Explanation:

H+ concentration of solutions is related to pH as -

[H+] = antilog(-pH)

For pH = 4,

[H+] = antilog(-4)

[H+] = 10^-4 M

For pH = 10,

[H+] = antilog(-10)

[H+] = 10^-10 M

When these 2 solutions are added, concentration of the mixture will be -

[H+] = √(10^-10 × 10^-4)

[H+] = √(10^-14)

[H+] = 10^-7 M

So now, pH of the mixture will be

pH = -log(10^-7)

pH = 7