If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right

Answers

Answer 1

Answer:

E. Q < K and reaction shifts right

Explanation:

Step 1: Write the balanced equation

A(s) + 3 B(l) ⇄ 2(aq) + D(aq)

Step 2: Calculate the reaction quotient (Q)

The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.

Q = [C]² × [D]

Q = 0.64² × 0.38

Q = 0.15

Step 3: Compare Q with K and determine in which direction will shift the reaction

Since Q < K, the reaction will shift to the right to attain the equilibrium.


Related Questions

Sometimes a nuclide is referenced by the name of the element followed by the:______
a. atomic number
b. mass number
c. electrical charge
d. none of the above

Answers

Answer:

The correct option is d

Explanation:

Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.

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.

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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https://brainly.com/question/6019799

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.

A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.

Answers

I’m pretty sure that they mostly design building structures!

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

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.

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

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.

Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent

Answers

Answer:

Lewis acid

Explanation:

In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.

If we look at the formation of PF6^-, the process is as follows;

PF5 + F^- -----> PF6^-

We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.

Hence in the formation of PF6^-, PF5 acts a Lewis acid.

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

A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
(1) Before the addition of any hydrobromic acid, the pH is___________.
(2) After adding 12.0 mL of hydrobromic acid, the pH is__________.
(3) At the titration midpoint, the pH is___________.
(4) At the equivalence point, the pH is________.
(5) After adding 45.1 mL of hydrobromic acid, the pH is_________.

Answers

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

what is radiologist​

Answers

Radiologists are medical doctors that treat injuries using medical imaging (radiology)

Answer:

a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.

Explanation:

The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate boiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC

Answers

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]

We can solve for the temperature as follows:

[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]

Thus, with the proper units, we obtain:

[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]

Hence, answer is approximately 100 °C.

Best regards.

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

Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene

Answers

Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .  

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

c)

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring  are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

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.

A student accidentally let some of the vapor escape the beaker. As a result of this error, will the mass of naphthalene you record be too high, too low, or unaffected? Why?

Answers

Answer:

too low

Explanation:

If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.

Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.

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]

When alkanes react with chlorine in the presence of ultraviolet light, chlorine atoms substitute for one or more alkane hydrogen atoms. What is the number of different chloroalkane compounds that can be formed by the reaction of C2H6 with chlorine?

Answers

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6

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.

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

what bsic difference is between NMR and MS spectroscopic techniques?​

Answers

Answer:

The Nuclear magnetic resonance is the process this technique does not use radiation.

The  ms is an sensitive technology can be a massive number and small sample of the blood.

Explanation:

The Nuclear magnetic resonance we look at the both side of that coin.

The technique provides that fatty acid composition and various including amino acids.

These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-

(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.

The MS is an extremely sensitive technology using a very small number of the blood.

(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume

MS can be fine mapping metabolic pathways to sign analytical strategy.

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

suppose you are titrating vinegar, which is an acetic acid solution

Answers

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

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.

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

For dinner you make a salad with lettuce, tomatoes, cheese, carrots, and
croutons. Your salad would be classified as a(n)
O A. compound
OB. element
OC. homogeneous mixture
D. heterogeneous mixture​

Answers

A heterogeneous mixture​

In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 176 kg/h. Determine the percentage of excess air used during this process.

Answers

Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

What is the osmolarity of a 0.20 M solution of KCI?
A) 0.40 Osmol
B) 0.30 Osmol C) 0.20 Osmol D) 0.80 Osmol
E) 0.10 Osmol

Answers

Answer:

Osmolarity of solution of KCI = 0.40 osmol

Explanation:

Given:

KCL ⇒ K⁺ + Cl⁻

Find:

Osmolarity of solution of KCI

When M = 0.20 M

Computation:

1 mole of KCL = 2 osmol

1 M of KCl = 2 Osmolarity

So,

Osmolarity of solution of KCI = 2 × 0.20

Osmolarity of solution of KCI = 0.40 osmol

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