What role does wind play in mechanical weathering?

Answer:

Warm

Explanation: because it would be less hot air population

Answer:

Explanation:

a)   the units of the constants 200 and 10 are as follows:

unit of 200 = unit of r / unit of D

= crystals/min× mm

= crystals / (min×mm)

unit of 10 = unit of r / unit of D^2

= crystals/min × mm²

= crystals / (min×mm²)

b) The objective here is to determine the crystal nucleation rate in crystals/s corresponding to a crystal diameter of 0.050 inch; T o do that ; let's first convert the inch to  mm

We all know that

1 inch = 25.4 mm

0.050 inch = 0.050 ×25.4 mm

= 1.27 mm

nucleation rate = 200×D - 10×D²

= 200×1.27 - 10×(1.27)²

=237.9 Crystals/min

=237.9/60 crystals/sec

= 3.96 crystals/sec

c) Derive a formula for r(crystals/s) in terms of D(inches). (See Example 2.6-1.) Check the formula using the result of Part b.

r(crystals/sec)=A D−B D² (D in inch)

unit of 200= crystals / (min×mm)

unit of 10=crystals / (min×mm² )  

A = 200 crystals / (min×mm) × 1/60 min/sec ×25.4 mm/inch

= 84.7 crystals/(sec-inch)

B = 10 crystals / (min×mm² ) × 1/60 min/sec ×25.4 mm/inch×25.4 mm/inch

=107.5 crystals/(sec-inch)

Answer:

Van't Hoff factor of KBr is 1.63

Explanation:

Freezing point depression due the addition of a solute follows the formula:

ΔT = Kf×m×i

Where ΔT is change in freezing point, Kf is freezing point depression constant of the solvent X, m is molality of the solution (mol / kg) and i is Van't Hoff factor.

Moles of 177g of alanine (Molar mass: 89.09g/mol) are:

177g × (1mol / 89.09g) = 1.99 moles. In 0.800kg:

1.99mol / 0.800kg = 2.49m

Replacing in freezing point depression formula:

5.9°C = Kf×2.49m×1

Alinine has a Van't Hoff factor of 1

The Kf of the solvent is:

2.37 °C/m

Molality of the 177.0g of KBr solution (Molar mass: 119g/mol) is:

177.0g × (1mol / 119g) = 1.487 moles / 0.800kg = 1.859m

And the freezing point depression formula is:

7.2°C = 2.37°C/m×1.859m×i

1.63 = i

Van't Hoff factor of KBr is 1.63

Answer:

[tex]V_2=9.97L[/tex]

Explanation:

Hello,

In this case, we can apply the Boyle's law in order to understand the pressure-volume relationship as an inversely proportional relationship:

[tex]P_1V_1=P_2V_2[/tex]

Thus, we can solve for the resulting volume at the second state as follows:

[tex]V_2=\frac{P_1V_1}{P_2}=\frac{5.0L*760torr}{389tor} \\\\V_2=9.97L[/tex]

Best regards.

Answer:

C

Explanation:

Following are the calculation to units of [tex]\bold{NO_{3}}[/tex]:

Given equation:

[tex]\bold{3Cu + 8HNO_3 \longrightarrow 3Cu(NO_3)2 + 2NO + 4H_2O}[/tex]

To find:

units of [tex]\bold{NO_{3}}[/tex]=?

Solution:

[tex]\bold{3Cu + 8HNO_3 \longrightarrow 3Cu(NO_3)2 + 2NO + 4H_2O}[/tex]

Therefore, the final answer is "Option C".

Learn more:

brainly.com/question/18492732

Answer:

Structural formulas are the graphical representation of chemical compunds and shows the chemical bonds between the atoms of a molecule.

structural formulas of following compunds is attached below:

2,4-dimethylhexane - C8H18, it will have single bonds between carbon atom and will have methyl group at position 2 and 4 carbon.

4-methyl-2-pentene - C6H12, it will have double bond at 2nd carbon and methyl group at 4th carbon.

4-chloro-7-methyl-2-nonyne - C10H17Cl, it will have triple bond at 2nd position, chloride group at 4th carbon and a methyl group at 7th carbon.

Answer:

The concentration of hydronium ions and the pH value is related by the equation: pH=-Log[H+]

Explanation:

We have two different concepts pH and concentration of hydronium ions. Lets start with the hydronium ion.

An hydronium ion is a species that is produced by an acid:

[tex]HA~->~H^+~+~A^-[/tex]

Aditionally, we can have the production of hydroxide ions. The subtances that have the capacity to produce this ions are called "bases":

[tex]BOH~->~B^+~+~OH^-[/tex]

Now we can continue "pH"

The pH is a scale that indicates if the substance is and acid (higher concentration of [tex]H^+[/tex]), neutral (equal amounts of [tex]H+[/tex] and [tex]OH^-[/tex]) or basic (higher amount of [tex]OH^-[/tex]).

Finally, the "pH" is calculated with the concentration of the hydronium ions (

[tex]H^+[/tex]), the letter "p" is "-Log", therefore:

[tex]pH=-Log[H^+] [/tex]

I hope it helps!

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Answer:

Gneiss is a foliated metamorphic rock

Explanation:

Gneiss  is a high grade metamorphic rock, meaning that it has been subjected to higher temperatures and pressures .

Answer:

4.67 kg

Explanation:

Given data

Step 1: Calculate the volume of the sheet

The volume of the sheet is equal to the product of its dimensions.

[tex]V = 75.0 cm \times 55.0 cm \times 0.10 cm = 413 cm^{3}[/tex]

Step 2: Calculate the mass of the sheet

The density (ρ) is equal to the mass divided by the volume.

[tex]\rho = \frac{m}{V} \\m = \rho \times V = \frac{11.3g}{cm^{3} } \times 413cm^{3} = 4.67 \times 10^{3} g = 4.67 kg[/tex]

Answer:

Explanation:

D

Answer: The temperature rise is [tex]0.53^0C[/tex]

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g   (1kg=1000g)

c = heat capacity of water = [tex]4.18J/g^0C[/tex]

Initial temperature  = [tex]T_i[/tex]

Final temperature = [tex]T_f[/tex]  

Change in temperature ,[tex]\Delta T=T_f-T_i=?[/tex]

Putting in the values, we get:

[tex]5280J=2400g\times 4.18J/g^0C\times \Delta T[/tex]

[tex]\Delta T=0.53^0C[/tex]

Thus the temperature rise is [tex]0.53^0C[/tex]

Answer: 0.580 C

Explanation: On Ck-12 I got it right sooo...

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Since there has been a rise in the reaction temperature, there has been an exothermic reaction.

The amount of heat energy released in the second step has been -132.54 kJ.

The reaction enthalpy per mole has been -1160.85 kJ/mol.

(a) To determine whether the reaction has been exothermic or endothermic, the heat absorbed or released has been calculated.

Since there has been a rise in the temperature of the solution with the combustion, the reaction has been termed as the exothermic reaction.

(b) Amount of heat released in second experiment:

In the bomb calorimeter:

[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]

q = mc[tex]\Delta[/tex]T

q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (36.99 - 10[tex]\rm ^\circ C[/tex] )

[tex]\rm q_w_a_t_e_r[/tex] = 169389.24 J.

q = C [tex]\Delta[/tex]T

q = c (36.99 - 10[tex]\rm ^\circ C[/tex] )

[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c

q = mass × heat of combustion of benzoic acid

q = 7.5 g × 26.454 kJ/g

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 198405 J

[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]

[tex]\rm q_b_o_m_b\;[/tex] = - ([tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_w_a_t_e_r[/tex])

[tex]\rm q_b_o_m_b\;[/tex] = - (198405 J + 169389.24 J )

[tex]\rm q_b_o_m_b\;[/tex] = 29015.76 J.

[tex]\rm q_b_o_m_b\;[/tex] = 26.99 [tex]\rm ^\circ C[/tex] × c

29015.76 J = 26.99 [tex]\rm ^\circ C[/tex] × c

c of bomb = 1075.05 J/[tex]\rm ^\circ C[/tex].

For the second reaction of combustion of ethanol:

[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]

q = mc[tex]\Delta[/tex]T

q = 1500 g × 4.184 J/g.[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )

[tex]\rm q_w_a_t_e_r[/tex] = 113156.28 J.

q = C [tex]\Delta[/tex]T

q = 1075.05 J/[tex]\rm ^\circ C[/tex] × (28.03 - 10[tex]\rm ^\circ C[/tex] )

[tex]\rm q_b_o_m_b\;[/tex] = 19383.15 J

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

Moles of ethanol = [tex]\rm \dfrac{5.26\;g}{46.07\;g/mol}[/tex]

Moles of ethanol = 0.11417 mol.

q = moles of ethanol × [tex]\Delta[/tex]H of reaction

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417  × [tex]\Delta[/tex]H of reaction

[tex]\rm q_r_e_a_c_t_i_o_n\;+\;q_b_o_m_b\;+\;q_w_a_t_e_r\;=\;0[/tex]

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - ([tex]\rm q_b_o_m_b\;[/tex] + [tex]\rm q_w_a_t_e_r[/tex])

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = - (19383.15 J +  113156.28 J)

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132539.43 J

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = -132.54 kJ.

The amount of heat energy released in the second step has been -132.54 kJ.

(c) The  reaction enthalpy per mole can be given as:

[tex]\rm q_r_e_a_c_t_i_o_n[/tex] = 0.11417 mol  × [tex]\Delta[/tex]H of reaction

-132.54 kJ = 0.11417 mol  × [tex]\Delta[/tex]H of reaction

[tex]\Delta[/tex]H of reaction = -1160.85 kJ/mol.

The reaction enthalpy per mole has been -1160.85 kJ/mol.

Since there has been a rise in the reaction temperature, there has been an exothermic reaction.

The amount of heat energy released in the second step has been -132.54 kJ.

The reaction enthalpy per mole has been -1160.85 kJ/mol.

For more information about the reaction in the bomb calorimeter, refer to the link:

https://brainly.com/question/14989357

Explanation:

when NaSO4 added, Ca2+ + SO42- = CaSO4

and Calcium sulphate is a white insoluble precipitate

when NaCO3 added, Fe2+ + CO32- = FeCO3

and it is a green precipitate

Answer:

Yes it is. 0.34 kg = 340 g (Kilograms and grams)

Explanation:

When we convert from kg to grams we multiply by 1000 because kilo mean 1000.

0.34*1000= 340

Answer:

Option A: It is likely that more rock candy will be formed in batch A.

Explanation:

The difference between batch A and batch B is that batch A uses temperature to dissolve the sugar, while the dissolution of the sugar in batch B is produced at room temperature.

When he use temperature (hot water) to dissolve the sugar he is increasing the solubility of the sugar in the water, so in batch A we will have more quantity of sugar dissolved than in the batch B. The cooling of the solution at room temperature favors the formation of bigger sugar crystals in the process of crystallization.

From all of the above, the correct predicted observation is the option A: It is likely that more rock candy will be formed in batch A, for the increase of the solubility by the use of hot water. Also, you say that Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization.

I hope it helps you!  

Answer:

Half-lives of first order reactions

[A]1/2[A]o=12=e−kt1/2.

ln0.5=−kt.

t1/2=ln2k≈0.693k.

Answer:

r= 3 inches

Explanation:

V= (4/3)*pi* r^3

113.04= (4/3)*3.14*r^3

113.04*(3/4)= 3.14*r^3

84.78 = 3.14*r^3

84.78/3.14 = r^3

27 = r^3 Take the cubed root of both sides.

r = 3 inches

Answer:

The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l)  ⇄ N₃⁻(aq) + H₃O⁺(aq)          

The pH of this solution is:

[tex] pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60 [/tex]

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq)  ⇄  HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

[tex] \eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles [/tex]

Now, the number of moles of HN₃ is:

[tex] \eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles [/tex]

Then, the pH of the buffer solution after the addition of HCl is:

[tex] pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43 [/tex]

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.    

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

Answer:

Dissolve benxioc acid and benzaldehyde in organic solvent. The two compounds are not miscible.with water. Put the two in separating funnel. Then use aqueous sodium bicarbonate to extract. Benzioc acid will be in aqueous layer as benzioate ion. Benzaldehyde remain insoluble and can be isolated.

Explanation:

Extractions are techniques use to separate desired compounds when mixed together. The mixture is brought in contact with solvent in which the sites substance is soluble and other is insoluble. Extractions use imissicible stages to separate substance from another.

Answer:

[tex]\mathbf{t_2 = 75.696 \ min}[/tex]

Explanation:

From the question:

The outer surface of a steel gear is to be hardened by increasing its carbon content

Given that :

Diffusion of heat temperature at [tex]T_1[/tex] 850 °C  = 1123 K

Diffusion time [tex]t_1[/tex] = 10 min

diffusion after the carbon concentration at a position [tex]x_1[/tex] ( 1.0 mm) below the surface =  0.90 wt%

Preexponential = 1.1 × 10⁻⁶ m²/s

Activation Energy [tex]Q_d[/tex] = 87400 J/mol

We are to determine the time [tex]t_2[/tex] at 650  °C (923 K) to achieve the same diffusion result as at 850 °C (1123 K) for [tex]t_1[/tex] = 10 min

Considering Fick's second law for the condition of Constant surface concentration; we have:

[tex]\frac{Cx-C_0}{C_s-C_0} = 1-erf(\frac{x}{2\sqrt{Dt} } )[/tex]   ------ equation (1)

where;

[tex]C_0 =[/tex] concentration of the diffusing solute atom before diffusion

[tex]C_s[/tex] = Constant surface concentration

[tex]C_x[/tex] = Concentration at depth x after time t

[tex]erf(\frac{x}{2\sqrt{Dt} } )[/tex] = Gaussian error function

At some desired specific  concentration of solute [tex]C_1[/tex] in an alloy ; the left side of the above equation (1) thus becomes constant ;

i.e [tex]\frac{Cx-C_0}{C_s-C_0} = \mathbf{ constant}[/tex]

Then ;  [tex]\frac{x}{2\sqrt{Dt} }[/tex] = constant

[tex]\frac{x^2}{Dt}[/tex] = constant

Dt = constant

Thus; [tex]D_1t_1 = D_2t_2[/tex]

Therefore, the time [tex]t_2[/tex] at 650°C([tex]T_2[/tex] = 923 K) required to produce the same diffusion on result as at 850°C ([tex]T_1[/tex] = 1123 K) for [tex]t_1[/tex] = 10 min is [tex]t_2 = \frac{D_1t_1}{D_2}[/tex]

We need to first determine the Diffusion coefficient at 1123 K and 923 K ( i.e [tex]D_1[/tex]  and  [tex]D_2[/tex])

At [tex]T_1[/tex] = 1123 K , Diffusion coefficient [tex]D_1[/tex] is calculated by the equation [tex]D_1 = D_0 exp ( - \frac{Q_d}{RT_1})[/tex]       (equation from temperature dependence of the diffusion coefficient)

[tex]D_1 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*1123} )[/tex]

[tex]D_1 = 9.462*10^{-11} \ m^2/s[/tex]

[tex]D_2 = D_0 exp ( - \frac{Q_d}{RT_2})[/tex]

[tex]D_2 = 1.1 * 10^{-6} \ exp ( - \frac{87,400}{8.314*923} )[/tex]

[tex]D_2 = 1.25*10^{-11} m^2/s[/tex]

[tex]t_2 = \frac{ 9.462*10^{-11}*10}{ 1.25*10^{-11} }[/tex]

[tex]\mathbf{t_2 = 75.696 \ min}[/tex]

Answer:

decreases

Explanation:

The higher the entropy, the more unpredictable and the more random a value is, this is because entropy describes disorder of a system. If things are less random, that means they are predictable, which means they are more ordered.

Answer:

In this reaction, [tex]6\; \rm mol[/tex] of [tex]\rm N_2[/tex]  will react with [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex].

Explanation:

In the balanced equation for this reaction, the ratio between the coefficient of [tex]\rm H_2[/tex] and [tex]\rm N_2[/tex] is [tex]3 : 1[/tex]. That is: [tex]n(\mathrm{H_2}) : n(\mathrm{N_2}) = 3:1[/tex]. That's the same as saying that for every one mole of [tex]\rm N_2[/tex] consumed, three moles of [tex]\rm H_2[/tex] will be consumed.

Rewrite this ratio as a fraction:

[tex]\displaystyle \frac{n(\mathrm{H_2})}{n(\mathrm{N_2})} = \frac{3}{1}[/tex].

Take the reciprocal of both sides to obtain:

[tex]\displaystyle \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})} = \frac{1}{3}[/tex].

It is given that [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex] was consumed; in other words, [tex]n(\mathrm{H_2}) = 18\; \rm mol[/tex]. The question is asking for [tex]n(\mathrm{N_2})[/tex], the number of moles of [tex]\rm N_2[/tex] required. Apply this ratio:

[tex]\begin{aligned}n(\mathrm{N_2}) &= n(\mathrm{H_2}) \cdot \frac{n(\mathrm{N_2})}{n(\mathrm{H_2})}\\ &= 18\; \rm mol \times \frac{1}{3} = 6\; \rm mol\end{aligned}[/tex].

Hence the conclusion: in this reaction, it will take (at least) [tex]6\; \rm mol[/tex] of [tex]\rm N_2[/tex] to react with [tex]18\; \rm mol[/tex] of [tex]\rm H_2[/tex].

Answer:

yes

Explanation: took quiz

Answer:

In effusion, a substance escapes through a tiny pinhole, or other hole whereas diffusion is the spreading out of a substance within a dispersing medium.

Effusion and diffusion are similar in terms of rates and speeds.

Explanation:

According to effusion, if there is a small hole in a container, a gas will leak from it. So, in this process a substance escapes through a tiny pinhole, or other hole.

According to diffusion, the two solutions will reach the lowest possible concentration of both, if they are combined and not mixed. So, diffusion is the spreading out of a substance within a dispersing medium.

Effusion and diffusion are similar in terms of rates and speeds.

Answer:

Explanation:

The objective of this  question is all about identifying the phenomena that holds true for the statement being said in each instance. Let; walk through them.

a) Whenever you see 9 or a multiple of 9 in the integration ratio, which group should you first consider for being responsible for that signal.

( C₄H₉ )Tert. Butyl group

b) Whenever you see a quartet and triplet on a spectrum, which group should you first consider for being responsible for those signals?

(CH₃CH₂)  ethyl group

c) Whenever you see septet and a doublet on a spectrum, which group should you first consider for being responsible for those signals?

(CH(CH₃)₂)  isoproply group

d) Whenever you see 6 or a multiple of 6 in the integration ratio, which group should you first consider for being responsible for that signal

(CH(CH₃)₂)  isoproply group

e)  Whenever you see 3 as the actual number of protons for a given signal, which group should you first consider for being responsible for that signal.

CH₃- methyl group

A perfect description showing  the explanation of each answers chosen is explained with an aid of diagram below.

Answer:

D. (HCIO4(AQ)

Explanation:

Answer:

Q = 21.896kJ

Explanation:

Q = ?

∇U = 21.39kJ

W = ?

W = P∇V

W = P (V2 - V1)

W = 0.276 × (1.876 - 0.04232)

W = 0.276 × 1.83368

W = 0.5060J

Q = ∇U + W

Q = 21.39 + 0.5060

Q = 21.896kJ

The energy change corresponds to the work done by the system

Answer:

[tex]\Delta E=21.34kJ[/tex]

Explanation:

Hello,

In this case, we should apply the first law of thermodynamics to compute the energy change:

[tex]\Delta E=Q-W[/tex]

Thus, with the given volume change we compute the corresponding work in kJ:

[tex]W=P\Delta V=0.276atm*(1.876L-0.0432L)*\frac{101.325kPa}{1atm}*\frac{1m^3}{1000L}=0.0513kJ[/tex]

Then, we compute the energy change:

[tex]\Delta E=21.39kJ-0.0512kJ\\\\\Delta E=21.34kJ[/tex]

Best regards.