The diagram shows two waves.
How do the frequencies of the waves compare?
Wave A has a lower frequency because it has a
smaller amplitude.
Wave A has a higher frequency because it has a
shorter wavelength.
The waves have the same frequency because they
have the same wavelength.
The waves have the same frequency because they
have the same amplitude.

The Diagram Shows Two Waves.How Do The Frequencies Of The Waves Compare?Wave A Has A Lower Frequency

Answers

Answer 1

Answer:

Wave A has a higher frequency because it has a shorter wavelength.

Explanation:

The frequency of a wave and the wave length are related by the following equation:

Velocity (v) = wave length (λ) x frequency (f)

v = λf

If we make frequency (f) the subject of the above equation, we will have:

f = v/λ

Let the velocity (v) be constant.

f = v/λ

f & 1/λ

From the equation above,

We can see that the frequency (f) is inversely proportional to the wavelength (λ).

This implies that a wave with a high frequency, will have a short wavelength and a wave with a short frequency will have a longer wavelength.

Now considering wave A and B in the diagram above,

Wave A will have a higher frequency because it has a shorter wavelength as explained above.

Answer 2

Answer:

it is the second option

Explanation:


Related Questions

D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match the words in the left column to the appropriate blanks in the sentences on the right. Fill in the blanks.
a ketone
carbon 3
carbon 2
carbon 1
an aldehyde
carbon 4
In D-glucose, there is__________ functional group, and the carbonyl group is at___________ when looking at the Fischer projection.
In D-tructose. there is functional group, and the carbonyl group is at when looking at______ the Fischer projection.

Answers

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

Arrange the following elements in order of decreasing first ionization energy: S, Ca, F, Rb, and Si.
Rank from largest to smallest. To rank items as equivalent, overlap them.

Answers

Answer:

The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.  

Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.  

Considering the definition of ionization energy,

Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.

You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.

In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.

In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.

Taking into account the above, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.  

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The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine

Answers

Answer:

An alkyl halide can undergo SN2 reaction with an amine

Explanation:

The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.

The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).

This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.

The detailed mechanism of this reaction has been attached to this answer.

g Increasing the number of unsaturations in a fatty acid ____________ the melting temperature of the fatty acid.

Answers

Answer:

Decreases

Explanation:

Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.

So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.


When the owners of some wells in Pallerla started using high-powered motors to
draw water from the wells, the owners of other wells noticed that their wells were
drying up. Discuss the possible solution to the problem solutions to the problem​

Answers

Answer:

The possible solution is to balance the rate of water removal from the well to the rate of natural recharge of the well from its underground aquifer.

Explanation:

A well is an excavation in the earth, made with the aim of extracting water from the aquifers. The water from a well can be drawn up by the means of a pump, containers, such as buckets, or by hand. Aquifers can also be recharged through a well.

Well draw down occurs when water from the well is drained faster than it is naturally recharged from the aquifer. This can be as a result of over pumping, extended drought, among other factors. The use of the high-powered motor in this case, for pumping, might be the possible cause of the well drying up. The situation might have resulted from the pump drawing out water from the well at a rate tat exceeds the rate at which it is recharged naturally, causing the well water to start drying up. There's also a possibility that the well is pumped indiscriminately, possibly leading to wastage of water.

The solution to this problem is to give the well a time duration for it to recharge itself. Then, the rate of recharges should be calculated and determined by an hydrologist. When all these is done, a pump with a motor power that does not exceed the calculated recharge rate should be used in place of the high-powered motor. Also, water usage should be brought to the minimum level to prevent unnecessary pumping due to excessive, wasteful use of water.

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en la siguiente reacción: CH3NH2 + O2 → CO2 + H2O + N2 Si reaccionan 0,5 mol de metil amina (CH3NH2) con 25,6 g de O2. Determine: a) Balancee la ecuación. (2 ptos) b) ¿Cuántos gramos de nitrógeno (N2) se pueden producir? (4 ptos) c) Si experimentalmente se obtuvieron 3,5 gramos de N2. Determine el porcentaje de rendimiento de la reacción. (4 ptos) Por favor es urgente!!!

Answers

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂    

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂              

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.  

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

[tex]n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles[/tex]      

[tex]n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles[/tex]

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

[tex]n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles[/tex]

[tex]m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g[/tex]

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

[tex] \% = \frac{R_{r}}{R_{T}}*100 [/tex]

Donde:

[tex]R_{r}[/tex]: es el rendimiento real

[tex]R_{T}[/tex]: es el rendimiento teórico

[tex]\% = \frac{3,5}{5,043}*100 = 69,4[/tex]

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!        

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL) with excess benzaldehyde and NaOH to produce 79.4 g of (1E,4E)-1,5-diphenylpenta-1,4-dien-3-one (234.29 g/mol). What is the percent yield of this student's experiment

Answers

Answer:

% yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Explanation:

given

volume of acetone= 43.8 mL

molar weight of acetone = 58.08 g/mol

density of acetone = 0.791 g/mL

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)

43.8 mL = 43.8mL × 0.791g/mL

= 34.6458g ≈34.65g

1 mole of acetone = 58.08g

∴34.65g = 34.65g/58.08g

= 0.60mol

molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol

mole = mass/ molar weight

mole = 79.4g/ 234.29g/mol

mole(n) = 0.3389mol ≈ 0.34mol

1 mole of acetone will produce 1 mole of the product

∴0.60mol of acetone will produce 0.60mol of the product

but we get 0.34mol of the product

∴ % yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of manganese(II) nitrate and sodium hydroxide are combined.

Answers

Answer:

Explanation:

Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ +  2Na NO₃

Converting into ions

Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻

Cancelling out common terms

Mn⁺ + 2 OH⁻ = Mn( OH)₂

this is net ionic equation required.

It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. single by absorbing a significant digit.

Answers

Answer:

495nm

Explanation:

The energy of a photon could be obtained by using:

E = hc / λ

Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.

The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:

242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.

Replacing in the equation:

E = hc / λ

4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ

λ = 4.946x10⁻⁷m

Is maximum wavelength  of light that could break a Cl-Cl bond.

Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:

4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =

495nm

Which of these groups of elements show the least electronegativity?

Answers

Explanation:

On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group. As a result, the most electronegative elements are found on the top right of the periodic table, while the least electronegative elements are found on the bottom left. The answer is alkali metals.

What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A? Express your answer using two significant figures.

Answers

Answer

mass of aluminum metal= 7 .0497g of Al

Explanation:

current = 21 A

time = 1 hour = 60 X 60 = 3600 s

quantity of electricity passed = current X time = 21X 3600 = 75600 C

Following the electrolysis the below reaction will occur :

Al3+ + 3e- --------> Al

therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al

so 1 C will produce 1/289500 moles of Al

so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al

now 1 mole of aluminium weighs = 27 g/mole

so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g

mass of aluminum metal= 7 .0497 g of Al

The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g

We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:

Current (I) = 21 A

Time(t) = 1 h = 60 × 60 = 3600 s

Quantity of electricity (Q) =?

Q = it

Q = 21 × 3600

Q = 75600 C

Finally, we shall determine the mass of the aluminum metal produced.

Al³⁺ + 3e —> Al

Recall:

1 mole of Al = 27 g

1 electron (e) = 96500 C

Thus,

3 electrons = 3 × 96500 = 289500 C

From the balanced equation above,

289500 C of electricity produced 27 g of Al.

Therefore,

75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al

Thus, the mass of the aluminum metal obtained is 7.05 g

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A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the decay rate for 1.000 mole of this nuclide. decays s-1

Answers

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

Which sample is most likely to experience the smallest temperature change upon observing 55KJ of heat? 

Answers

Answer:

100 g of water: specific heat of water 4.18 J/g°C

Explanation:

To know the correct answer to the question, we shall determine the temperature change in each case.

For 100 g of water:

Mass (M) = 100 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 4.18 x ΔT

Divide both side by 100 x 4.18

ΔT = 55000/ (100 x 4.18)

ΔT = 131.6 °C

Therefore the temperature change is 131.6 °C

For 50 g of water:

Mass (M) = 50 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 4.18 x ΔT

Divide both side by 50 x 4.18

ΔT = 55000/ (50 x 4.18)

ΔT = 263.2 °C

Therefore the temperature change is 263.2 °C

For 50 g of lead:

Mass (M) = 50 g

Specific heat capacity (C) = 0.128 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 0.128 x ΔT

Divide both side by 50 x 0.128

ΔT = 55000/ (50 x 0.128)

ΔT = 8593.8 °C

Therefore the temperature change is 8593.8 °C.

For 100 g of iron:

Mass (M) = 100 g

Specific heat capacity (C) = 0.449 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 0.449 x ΔT

Divide both side by 100 x 0.449

ΔT = 55000/ (100 x 0.449)

ΔT = 1224.9 °C

Therefore the temperature change is 1224.9 °C.

The table below gives the summary of the temperature change of each substance:

Mass >>> Substance >> Temp. Change

100 g >>> Water >>>>>> 131.6 °C

50 g >>>> Water >>>>>> 263.2 °C

50 g >>>> Lead >>>>>>> 8593.8 °C

100 g >>> Iron >>>>>>>> 1224.9 °C

From the table given above we can see that 100 g of water has the smallest temperature change.

If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the course of radioactive decay, what is the mass number of the stable daughter product?

Answers

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH

Answers

Answer:

pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

Explanation:

pKa acetic acid, CH3COOH = 4.7

It is possible to determine pH of a buffer using H-H equation:

pH = pka + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH3COONa] / [CH3COOH]

As you want a pH 4.7 buffer:

4.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = [CH3COONa] / [CH3COOH]

That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:

5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

For pH 5.7:

5.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = log [CH3COONa] / [CH3COOH]

10 = [CH3COONa] / [CH3COOH] (1)

That means you need 10 times [CH3COONa] over [CH3COOH]

And as you know:

10mL=  [CH3COONa] + [CH3COOH] (2)

Replacing (1) in (2):

10 = 10mL + [CH3COOH] / [CH3COOH]

10[CH3COOH] = 10mL + [CH3COOH]

11[CH3COOH] = 10mL

[CH3COOH] = 0.91mL

And [CH3COONa] = 10mL - 0.91mL =

[CH3COONa] = 9.09mL

That is:

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

The volumes according to the pH are as follows:

(i) 5mL of 0.10 M CHCOOH and 5mL 0.10 M CHCOONa for pH 4.7

(ii) 0.91mL of 0.10 M CHCOOH and 9.09mL 0.10 M CHCOONa pH 5.7

Calculating the volume of chemicals needed:

Given that pKa of acetic acid, CH₃COOH = 4.7

The pH of a buffer using the H-H equation is given by:

pH = pKa + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH₃COONa] / [ CH₃COOH]

4.7 = 4.7 + log [CH₃COONa] / [ CH₃COOH]

0 = log [CH₃COONa] / [ CH₃COOH]

takin antilog on both sides of the equation we get:

1 = [CHCOONa] / [CHCOOH]

It implies that the same amount of both species is needed to make the pH 4.7 buffer.

So,

5mL of 0.10 M CHCOOH and 5mL 0.10 M CHCOONa makes a buffer of pH 4.7

Similarly:

5.7 = 4.7 + log [CH₃COONa] / [CH₃COOH]

1 = log [CH₃COONa] / [CH₃COOH]

takin antilog on both sides of the equation we get:

10 = [CHCOONa] / [CHCOOH]

10[CHCOOH] = [CHCOONa]

It implies that we need 10 times [CH₃COONa] as much of [CH₃COOH]

We have to prepare 10 mL of buffer, so:

10mL=  [CH₃COONa] + [CH₃COOH]

10mL = 11[CH₃COOH]

[CH₃COOH] = 0.91mL

So, [CH₃COONa] = 10mL - 0.91mL

[CHCOONa] = 9.09mL

Therefore,

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa is required to make a buffer of pH 5.7

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The decomposition of H2O2 is first order in H2O2 and the rate constant for this reaction is 1.63 x 10-4 s-1. How long will it take for [H2O2] to fall from 0.95 M to 0.33 M?

Answers

Answer:

It will take 6486.92 minutes  for [H2O2] to fall from 0.95 M to 0.33 M

Explanation:

The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.

In first order reaction;

[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]

where;

a = concentration at time t

[tex]a_o[/tex] = initial concentration

and k = constant.

[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]

[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]

[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]

t = 6486.92 minutes

Consider Zn + 2HCl → ZnCl2 + H2 (g). If 0.30 mol Zn is added to HCl, how many mol H2 are produced?

Answers

Answer:

0.3 mol

Explanation:

Assuming HCl is in excess and Zn is the limiting reagent,

from the balanced equation, we can see the mole ratio of Zn:H2 = 1:1,

which means, each mole of zinc reacted gives 1 mole of H2.

So, if 0.30 mol Zn is added, the no. of moles of H2 produced will also be 0.3 mol, since the ratio is 1:1.

Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm

please.

Answers

Answer:

3.00 L

Explanation:

Convert the pressure to Pascals.

P = 82 atm × (101325 Pa/atm)

P = 8,308,650 Pa

Convert temperature to Kelvins.

T = 27°C + 273

T = 300 K

Use ideal gas law:

PV = nRT

(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)

V = 0.00300 m³

If desired, convert to liters.

V = (0.00300 m³) (1000 L/m³)

V = 3.00 L

Answer:

[tex]\large \boxed{\text{3.0 L}}[/tex]

Explanation:

[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]

The amount of space an object takes up is called _____. gravity weight mass volume

Answers

Volume is the amount of space an object takes up

A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution

Answers

Answer:

0.0913 M

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, we obtained the following:

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 2

Data obtained from the question include:

Molarity of base, NaOH (Mb) = 0.0990 M

Volume of base, NaOH (Vb) = 20.52 mL

Volume of acid, H2C3H2O4 (Va) = 11.13 mL

Molarity of acid, H2C3H2O4 (Ma) =..?

The molarity of the acid, H2C3H2O4 can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 11.13 / 0.0990 x 20.52 = 1/2

Cross multiply

Ma x 11.13 x 2 = 0.0990 x 20.52 x 1

Divide both side by 11.13 x 2

Ma = (0.0990 x 20.52)/ (11.13 x 2)

Ma = 0.0913 M

Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M

What element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)​

Answers

Answer:

Option A

Explanation:

Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.

Answer:

A. Silicon (Si)

Explanation:

Silicon (Si) is primarily used as a semiconductor material to make electronic chips.

Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid.

Strong intermolecular forces

Weak intermolecular forces

Answers

Answer:

Strong intermolecular forces:  an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point.

Weak intermolecular forces: a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point.

Explanation:

Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces inclde dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.

The strength of the intermolecular forces in a liquid usually affects the various properties of the liquid such as viscosity, surface tension, vapour pressure and boiling point.

Strong intermolecular forces in a liquid results in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.

Weak intermolecular forces in a liquid results in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point of that liquid.

Strong intermolecular force is defined as the increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point while  weak intermolecular forces define as the decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in boiling point.

Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces include as follows:-

Dispersion forcesDipole-dipole interactionsHydrogen bondingion-dipole forces.

Strong intermolecular forces in a liquid result in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.

Weak intermolecular forces in a liquid result in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in the boiling point of that liquid.

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A vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West

Answers

Explanation:

Speed = 2345 ÷ 35 = 67m/s

Which of the following best describes hydrocarbons? a. Alkanes in which a hydrogen atom is replaced by a hydroxyl group b. Binary compounds of carbon and hydrogen c. Organic compounds containing water and carbon d. Covalently bonded carbon compounds which have intermolecular force attractions to hydrogen compounds e. Compounds which are formed by the reaction of a naturally occurring carbon-containing substance and water

Answers

Answer:

b. Binary compounds of carbon and hydrogen

Explanation:

Before proceeding, Hydrocarbons refers to organic chemical compounds composed exclusively of hydrogen and carbon atoms. This means the only elements present in an hydrocarbon are;

- Carbon

- Hydrogen

Looking through the options;

- Option A: This is wrong because the hydroxyl group contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option B: This is correct. Binary compounds refers to compounds with just two elements.

- option C: This is wrong because water contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option D: Carbon atoms can contain other elements so this option is wrong.

- option E: This also wrong because we had already gotten the correct option.

what are the similarities between amorphous solid and crystalline solid

Answers

Answer:

solid dont know

Explanation:

so sorry ask another

What is titration? Question 1 options: The process of quickly adding one solution to another until a solid is formed. The process of slowly adding one solution to another until the reaction between the two is complete. The process of mixing equal volumes of two solutions to observe the reaction between the two. The process of combining two solids until the reaction between the two is complete.

Answers

Answer:

The process of slowly adding one solution to another until the reaction between the two is complete.

Explanation:

When you perform a titration, you are slowly adding one solution of a known concentration called a titrant to a known volume of another solution of an unknown concentration until the reaction reaches neutralization, in which the reaction is no longer taking place. This is often indicated by a color change.

Hope that helps.

1. In the addition of HBr to conjugated dienes, is the product which results from 1,2-addition or that which results from 1,4-addition the product of kinetic control?
A. From 1,2-addition
B. From 1,4-addition
2. Which of the following is the strongest acid?
A. CH3CH20H
B. CHзOCH3
C. CH3CH
D. CH3COCH3
E. CH3COH

Answers

Answer:

The answer to this question can be defined as follows:

In question 1, the answer is "Option A".

In question 2, the answer is "[tex]\bold{CH_3COOH}[/tex]".

Explanation:

In the second question, there is mistype error in the choices so the correct answer to this question can be defined as follows:

The product From 1,2-addition as its consequence of 1,4-addition is the result of kinetic regulation by HBr in conjugated dienes.The chemical name of the [tex]CH_3COOH[/tex] is the acetic acid, it is one of the carboxylic acids quite basic. It is a major chemical production factor for use as disposable soft drinks, movies or wood glue, polyethylene terephthalate, and many plastics, fibers, and fabrics. It is also used in the storage of the water and soft drinks in the bottles.

A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas

Answers

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

[tex]R_g = 11.1 \ min[/tex]

Effusion rate of [tex]H_2[/tex],

[tex]R_h = 2.42 \ min[/tex]

Molar mass of hydrogen,

[tex]M_h = 1\times 2[/tex]

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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The cell potential for an electrochemical cell with a Zn, Zn2 half-cell and an Al, Al3 half-cell is _____ V. Enter your answer to the hundredths place and do not leave out a leading zero, if it is needed.

Answers

Answer:

0.900 V

Explanation:

Oxidation half cell;

2Al(s) -----> 2Al^3+(aq) + 6e

Reduction half equation;

3Zn^2+(aq) + 6e ----> 3Zn(s)

E°anode = -1.66V

E°cathode= -0.76 V

E°cell= E°cathode - E°anode

E°cell= -0.76-(-1.66)

E°cell= 0.900 V

Qualitatively estimate the relative melting points for each of the solids, and rank them in decreasing order.
Rank from highest to lowest melting point. To rank items as equivalent, overlap them.
sodium chloride
graphite
solid ammonia

Answers

Answer:

Graphite> sodium chloride> solid ammonia

Explanation:

Melting points of solids has a lot to do with the nature of intermolecular forces in the solid. A substance melts when the intermolecular forces holding the crystal lattice has been overcome such that that the crystal structure of the solid just collapses.

Graphite consists of covalently bonded layers of carbon atom which form a giant lattice. The melting point of graphite is very high because of the fact that the strong covalent bonds that hold the carbon atoms together in the layers require a lot of heat energy to break. Grapoghite melts at about 3600°C

Sodium chloride is an ionic compound that melts at about 801°C. The lattice is composed of alternate sodium and chloride ions.

Solid ammonia is held together by much weaker intermolecular interaction hence it has a melting point of about −77.73 °C.

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