does a periodic table indicate the occurrence of elements​

Answer:

potassium is likely to be harder

Explanation:

because whenan atom loss electron they can form particle ion acquare stable a arrangement that loos electron can transfer to another which that also to acquaire stable

A) 4

pH=-log[H+]

pH=-log(10^-4)

B) Weaker

pH of orange juice=4

pH of coffee=5

An acid with pH of 4 is stronger than a pH of 5

C) 7, neither/neutral

pH of water=7

Answer: The percent yield for the [tex]NaBr[/tex] is, 86.7 %

Explanation : Given,

Moles of [tex]FeBr_3[/tex] = 2.36 mol

Moles of [tex]NaBr[/tex] = 6.14 mol

First we have to calculate the moles of [tex]NaBr[/tex]

The balanced chemical equation is:

[tex]2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr[/tex]

From the reaction, we conclude that

As, 2 moles of [tex]FeBr_3[/tex] react to give 6 moles of [tex]NaBr[/tex]

So, 2.36 moles of [tex]FeBr_3[/tex] react to give [tex]\frac{6}{2}\times 2.36=7.08[/tex] mole of [tex]NaBr[/tex]

Now we have to calculate the percent yield for the [tex]NaBr[/tex].

[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

[tex]\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%[/tex]

Therefore, the percent yield for the [tex]NaBr[/tex] is, 86.7 %

Answer:

2.67761514 moles

Explanation:

0.075kg equates to 75g.

This means that the car emits 75g of CO per km.

dividing 75g by the molar mass of CO (28.01g/mol) we get 2.67761514 moles

Answer:

i just answered this and its a

Explanation:

Answer:

231.735 g/mol

Explanation:

Answer: The volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is LiOH.

We are given:

[tex]n_1=1\\M_1=0.10mol/L\\V_1=?mL\\n_2=1\\M_2=0.15mol/L\\V_2=10mL[/tex]

Putting values in above equation, we get:

[tex]1\times 0.10\times V_1=1\times 0.15\times 10\\\\V_1=15mL[/tex]

Thus the volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml

Answer:

They all have the same number of molecules or atoms which is 3.011*10²⁴ molecules or atoms.

5 moles of O = 3.011*10²⁴ molecules or atoms

5 moles of N = 3.011*10²⁴ molecules or atoms

5 moles of MgCl2 = 3.011*10²⁴ molecules or atoms

5 moles of C2H3NO = 3.011*10²⁴ molecules or atoms.

Explanation:

a)

5 moles of O = ?

1 mole of any substance is equal to Avogadro's number which is equal to 6.022×10²³molecules or atoms

From the question above,

5 moles of O = 5 × 6.022*10²³ = 3.011×10²⁴atoms or molecules.

b)

5 moles of N

From the same principle or fundamentally stated fact above,

1 mole of any substance = 6.022*10²³molecules or atom

5 moles of N = 5 × 6.024

2*10²³ = 3.011*10²⁴atoms or molecules.

c)

5 moles of MgCl₂

Same principle we used in a and b,

1 mole of any substance = 6.022*10²³ atoms or molecules

5 moles of MgCl2 = 5.0×6.022*10²³ = 3.011*10²⁴ atoms or molecules.

d)

5 moles of C₂H₃NO = ?

1 mole of C₂H₃NO = 6.022*10²³molecules or atoms

5 moles = x molecules or atoms

x = 5 × 6.024

2*10²³ = 3.011*10²⁴ atoms or molecules

What this proves is that no matter the compound, molecule or element, as long as they have equal amount of moles, they'll have the same number of atoms, molecules or particles.

Answer:

a. 3.011x10²⁴ atoms of O in 5 moles.

b. 3.011x10²⁴ atoms of N in 5 moles.

c. 3.011x10²⁴ atoms of MgCl₂ in 5 moles.

d. 3.011x10²⁴ atoms of C₂H₃NO in 5 moles

Explanation:

A mole is defined as the number of atoms in exactly 12.000g of ¹²C.

This number is equal to 6.022x10²³. That means 1 mole is equal to 6.022x10²³.

Thus:

a. O . 5 moles of oxygen (An atom) are:

5 moles O ₓ (6.022x10²³ atoms of O / mole) = 3.011x10²⁴ atoms of O in 5 moles

b. N . Also, 5 moles of nitrogen are:

5 moles N ₓ (6.022x10²³ atoms of N / mole) = 3.011x10²⁴ atoms of N in 5 moles

c. MgCl₂ . Magnesium chloride is a molecule. Again, 1 mole of MgCl₂ contains 6.022x10²³ molecules and 5 moles are:

5 moles MgCl₂ ₓ (6.022x10²³ molecules of MgCl₂ / mole) = 3.011x10²⁴ atoms of MgCl₂ in 5 moles

d. C₂H₃NO. 5 moles of C₂H₃NO are:

5 moles  C₂H₃NO ₓ (6.022x10²³ molecules of  C₂H₃NO / mole) = 3.011x10²⁴ atoms of C₂H₃NO in 5 moles

As you can see, number of molecules of 1 mole doesn't depend on the nature of the substance.

Answer: a)  Sn 2+  

Explanation:A reducing agent is a substance  that can be oxidized when it loses electrons while An oxidizing agent is a substance that is reduced when it gains   electrons.

 Considering  Cu 2+,   Fe 3+, Al 3+, Here,  Cu 2+ can  be reduce to its Copper metal, same for  Al 3+ to Aluminium  metal  and Fe 3+  to Fe 2+ ion  and will not be oxidized further since they are in their most oxidized states. The three ions are acting as only oxidizing agents.

while Sn 2+ can be oxidized to become Sn 4+ and also  be reduced to become Sn metal  acting as a reducing agent and an oxidizing agent.

Answer:

A.Scientists use genes to explain why evolution happens.

Explanation:

:)

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

Answer:

Octane number, measure of the ability of a fuel to resist knocking when ignited in a mixture with air in the cylinder of an internal-combustion engine.

Octane number is also known as octane rating. Octane numbers are based on a scale on which isooctane is 100 (minimal knock) and heptane is 0 (bad knock). The higher the octane number, the more compression required for fuel ignition. Fuels with high octane numbers are used in high performance gasoline engines.

(Hope this helps) Sky

Answer:

342.14g/mol.

Explanation:

The molar mass of Al2 (SO4)3 is 342.15g/mole.

Reasons that one aluminum atom=26.98. Therefore two aluminum atoms:=2(26.98 g/mole) =53.96 g/mole.

Mass of SO4=96.06g/mol.

Therefore molar mass of Al2(SO4)3= 53.96+(96.06)×3=342.14g/mol

Answer:

342.17 g/mol

Explanation:

Al= 26.98    

S= 32.07

O= 16.00

To calculate that you would have

2(26.98) since there are two Als

and then you have to distribute the 3 to the S and O so...

3S and 12O

which is

3(32.07) + 12(16.00)

which means you add

2(26.98) + 3(32.07) + 12(16.00)

=53.96 + 96.21 + 192

=342.17

there you Go! the final answer is 342.17 g/mol.

Answer:

For evaporation you need a warm, preferably humid temperature. The rate of evaporation increases with an increase in temperature. A windy climate is best, as wind helps to remove the evaporated water vapour, and therefore creating a better scope for evaporation to continue. The speed of wind is important for evaporation because the wind pulls in dry air, increasing the rate of evaporation.

Short Answer- Hot and humid temperature, lots of wind speed.

Answer:

Explanation:

1 Evaporation is the phase change of a liquid to a gas. There are three very important requirements for evaporation to take place, 1) available energy, 2) available water, and 3) a vertical moisture gradient. Approximately 600 calories of heat must be added to a gram of water for it to evaporate into the air.

2. Evaporation is the process of a substance in a liquid state changing to a gaseous state due to an increase in temperature and/or pressure. It is a fundamental part of the water cycle and is constantly occurring throughout nature.

3.Water evaporates faster if the temperature is higher, the air is dry, and if there's wind. The same is true outside in the natural environment. Evaporation rates are generally higher in hot, dry and windy climates.

                                thank u

Answer:

Industrial activity increases the emission of greenhouse gases in the atmosphere, especially carbon dioxide (CO2), and it negatively affects climate by increasing global warming.

Two actions to face the global warming:

1- To substitute the use of fossil fuels by alternative clean energy sources such as, for example, hydroelectric, geothermal, solar and wind energy resources.  

2- The government's policies decided to develop electric cars and to stimulate healthy habits such as walking instead of the use of conventional fossil-fueled transport modes.

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ

ΔH° of the reaction is for the production of SNG -747.54kJ

According to Hess’ law of constant summation, the standard reaction enthalpy is independent of the pathway or number of the reaction steps taken for a multistep reaction, rather it is the sum of standard enthalpies of intermediate reactions involved at the same temperature.

Based on Hess law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

From the given reactions:

The sum of Reaction 4 × 4 + Reaction 5 - Reaction 1 × 4 gives the reaction below:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = {-74.81 kJ × 4} - 890.3 kJ {- 4 ×-110.5}

ΔH° = -747.54kJ

Therefore, ΔH° of the reaction is for the production of SNG -747.54kJ

Learn more about enthalpy and Hess' law at: https://brainly.com/question/9328637

Answer: Thus the yield of uranium from 2.50 kg [tex]U_3O_8[/tex] is 2.12 kg

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex](6.023\times 10^{23})[/tex] of particles.

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}[/tex]

moles of [tex]U_3O_8=\frac{2.50\times 1000g}{842g/mol}=2.97mol[/tex]    (1kg=1000g)

As 1 mole of [tex]U_3O_8[/tex] contains = 3 moles of U

2.97 mole of [tex]U_3O_8[/tex] contains = [tex]\frac{3}{1}\times 2.97=8.91moles[/tex] moles of U

Mass of Uranium=[tex]moles\times {\text {Molar mass}}=8.91mol\times 238g/mol=2120g=2.12kg[/tex]  

 ( 1kg=1000g)

Thus the yield of uranium from 2.50 kg [tex]U_3O_8[/tex] is 2.12 kg

Answer:

Test tubes A and B turn a darker blue color.

Explanation:

Based on the information provided in the question it can be said that they should have noticed that Tube A and Tube B turned a dark blue color. This is because the starch turns the solution in the test tubes into a dark blue color due to the negative reaction caused by the conversion of starch present in salivary amylase

Answer:

D

Explanation:

Scientists always perform tests and observe and measure in the physical world to prove their points or answer their questions.

Answer:

El reactivo en exceso es hidrógeno

97.12g NH₃ son formados

Explanation:

Basados en la reacción:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

El hidrógeno pasa de estado de oxidación 0 a estado de oxidación +1. Al perder un electrón se oxida y es el agente reductor.

El nitrógeno pasa de estado 0 estado -3. Al ganar 3 electrones se reduce y es el agente oxidante.

100g de N₂ son (Peso molecular: 28g/mol):

100g × (1mol / 28g) = 3.57 moles de N₂

Y 25g de H₂ son (Peso molecular: 2g/mol):

25g × (1mol / 2g) = 12.5 moles de H₂

Como 3 moles de hidrógeno reaccionan por mol de nitrógeno, las moles de nitrógeno que se necesitan para hacer reaccionar completamente 12.5 moles de hidrógeno son:

12.5 moles H₂× (1 mol N₂ / 3 moles H₂) = 4.17 moles de nitrógeno.

Como hay 3.57 moles de nitrógeno, el reactivo en exceso es hidrógeno.

Como el reactivo limitante es el nitrógeno y 1 mol de nitrógeno produce 2 moles de amoniaco, las moles de amoniaco son:

3.57 moles de N₂ × (2 moles NH₃ / 1 mol N₂) = 7.14 moles de NH₃

La masa producida idealmente es:

7.14 mol NH₃ ₓ (17g/mol) = 121.4 g de NH₃. Como la eficiencia del proceso es del 80%:

121.4 g NH₃ × 80% = 97.12g NH₃ son formados

Answer:

Excess reactant: H₂

Mass of produced ammonia,  97.1 g

Explanation:

Identify the reaction:

N₂ + 3H₂  → 2NH₃

We identify the reducing agent and the  oxidizing agent:

N₂ changed the oxidation state from 0 to -3. This is the reduction, so it is the  oxidizing agent. By the way the H₂ is the reducing agent.

We convert the mass to moles:

100 g / 28 g/mol = 3.57 moles of N₂

25 g / 2 g/mol = 12.5 moles of H₂

Ratio is 1:3. For 1 mol of nitrogen, we need 3 moles of hydrogen

Then, 3.57 moles of N₂ would need (3.57 . 3) / 1 = 10.7 moles

We have 12.5 moles of H₂, so the hydrogen is the excess reactant and the nitrogen is the limiting.

To produce ammonia, the reaction needs 1 mol of N₂, that can produce 2 moles of product

3.57 moles of N₂ will produce (3.57 . 2) / 1 = 7.14 moles of NH₃

As yield reaction is 80%, we will produce 7.14 mol . 0.80 = 5.71 moles

We convert the moles to mass: 5.71 mol . 17 g / 1mol = 97.1 g

Answer:

The most important use of liquid crystals is in displays because the molecules of a liquid crystal can control the amount, color, and direction of vibration of the light that passes through them. hope that helps love!

Answer:

mass=279grams

Explanation:

GIVEN DATA

number of atoms=3.01×10^23

avogadro's number=6.022×10^22

molar mass of iron=56g/moles

TO FIND

mass in gram of iron=?

SOLUTION

by using the formula

mass in gram=(number of atoms÷avogdro's number)×molar mass

mass=(3.01×10^23÷6.022×10^23)×56

mass=0.499×56

mass=27.9grams=28 grams

Answer:

Rate of hydrogen formation is 0.05 mole per second

Explanation:

Firstly, we write the equation of reaction.

When alkali earth metals react with dilute mineral acid, the reaction is vigorous with the production or evolution of hydrogen gas as a result of the displacement of the hydrogen from the acid by the metal. This is one of the basic reactions of mineral acids

Ca + H2SO4 ——> CaSO4 + H2

Looking at the reaction, 1 mole of calcium gave 1 mole of the hydrogen gas

What we do now is to calculate the number of moles of calcium produced by 20g of Ca

Mathematically;

number of moles = mass/atomic mass

number of moles of calcium is thus

20/40 = 0.5 moles

Now, if 1 mole of calcium produced 1 mole of the gas

Definitely, 0.5 mole of calcium will produce 0.5 mole of the gas

So the rate of gas formation would be 0.5/10 = 0.05 mole/second

Answer:

See explanation below

Explanation:

You forgot to put the picture to do so. In this case, I manage to find one, and I hope is the one you are looking for. If not, then post it again and I'll gladly help you out again.

According to the picture with the answer, we have a cyclohexane with 4 methyl groups there. Two of them are facing towards the molecule with a darker bond. This means that the alkyl bromide, should have a bromine in one of the bonds, and in order to produce an E2 reaction, this bromine should be facing in the opposite direction of the methyl groups which are facing towards. This is because an E2 reaction occurs with the less steric hindrance in the molecule. If the bromine is in the same direction as the methyl group, it will cause a lot more of work to do a reaction, and therefore, an E2 reaction. I will promote instead a E1 or a sustitution product.

Therefore the alkyl bromide should be like the one in the picture 2.

Answer:

1. 0.56 mole of electron.

2. 0.28 mole of Co .

3. 16.52g of Co.

Explanation:

The following data were obtained from the question:

Current (I) = 15A

Time (t) = 1 hr = 1 x 60 x 60 = 3600secs

Next, we shall determine the quantity of electricity, Q flowing in electrolyte. This is illustrated below:

Quantity of electricity (Q) = current (I) x Time

Q = It

Q = 15 x 3600

Q = 54000C.

1. Determination of the moles of electrons used to electrolyzed the solution. This is illustrated below:

96500C = 1 mole of electron

Therefore, 54000C = 54000/96500 = 0.56 mole of electron.

Therefore, 0.56 mole of electron was used to electrolyze the solution.

2. Determination of the number of mole of Co produced at the cathode. This is illustrated below:

At the cathode:

Co2+ + 2e —> Co

From the balanced equation above,

2 moles of electron produce 1 mole of Co.

Therefore, 0.56 mole of electron will produce = 0.56/2 = 0.28 mole of Co.

Therefore, 0.28 mole of Co is produced at the cathode.

3. Determination of the mass of Co produced at the cathode. This can be achieved by doing the following:

Molar mass of Co = 59g/mol

Number of mole of Co = 0.28 mole

Mass of Co =..?

Mass = mole x molar mass

Mass of Co = 0.28 x 59

Mass of Co = 16.52g

Therefore, 16.52g of Co is produced at the cathode.

Answer: orbitals supernumerary

Explanation:

Answer:

both S orbital and p orbitals

Explanation:

took the exam got the question right.

Answer:

Radium-226 is a radioactive decay product in the uranium-238 decay series and is the precursor of radon-222. Radium-228 is a radioactive decay product in the thorium-232 decay series. Both isotopes give rise to many additional short-lived radionuclides, resulting in a wide spectrum of alpha, beta and gamma radiations.