CHEMISTRY COLLEGE

What energy transfer happens when wood is burning?

Answers

Answer 1

Answer:

Answer:

Mechanical to Heat

explanation:

The wood itself can make mechanical energy but when it's on fire it makes heat energy

Answer 2

Answer:

Answer: Chemical to heat and light

Explanation: The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs and produces heat and light.

Related Questions

In basic enzyme kinetics, a large Km indicates that the substrate binds A. Permanently B. Transiently C. Covalently D. with low specificity E. Weakly
What you understand by macrocylic effect in coordination chemistry
Metal vapor deposition is a process used to deposit a very thin layer of metal on a target substrate. A researcher puts a glass slide in the metal vapor deposition chamber and coats the slide with copper. After deposition, the glass slide had increased in mass by 2.26 milligrams. Approximately how many copper atoms were deposited on the glass slide
38. Consider the following equilibrium:2CO(g) + O2(g) =2CO2Keg=4.0 x 10-10What is the value of Key for 2CO2(g) + 2COR + O2g) ?
Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)$

$\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)$

$\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)$

$\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)$

$Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-(d[CH_4])/(dt)$

$\text{Rate of disappearance of }H_2O=-(d[H_2O])/(dt)$

$\text{Rate of formation of }CO=+(d[CO])/(dt)$

$\text{Rate of formation of }H_2=+(1)/(3)(d[H_2])/(dt)$

The rate of reaction expression is:

$\text{Rate of reaction}=-(d[CH_4])/(dt)=-(d[H_2O])/(dt)=+(d[CO])/(dt)=+(1)/(3)(d[H_2])/(dt)$

As we are given that:

$+(d[CO])/(dt)=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+(1)/(3)(d[H_2])/(dt)=+(d[CO])/(dt)$

$+(1)/(3)(d[H_2])/(dt)=0.35M/s$

$(d[H_2])/(dt)=3* 0.35M/s$

$(d[H_2])/(dt)=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

The sun warming the surface of a rock is
conduction
convection
radiation

Answers

Radiation..................

Radiation is your answer...

Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of potassium sulfide and chromium(II) nitrate are combined. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed leave it blank. 2Cr^3+ + 3S^2- (aq) + + rightarrow Cr_2S_4 (s) +

Answers

Answer:

S²⁻(aq) + Cr²⁺(aq) ⇄ CrS(s)

Explanation:

The molecular equation includes all the species in the molecular form. Usually, it is useful to write this first to balance the equation. This is a double displacement reaction.

K₂S(aq) + Cr(NO₃)₂(aq) ⇄ 2 KNO₃(aq) + CrS(s)

The full ionic equation includes all ions and the species that no dot dissociate in water.

2 K⁺(aq) + S²⁻(aq) + Cr²⁺(aq) + 2 NO₃⁻(aq) ⇄ 2 K⁺(aq) + 2 NO₃⁻(aq) + CrS(s)

The net ionic equation includes only those ions that participate in the reaction and the species that do not dissociate in water.

S²⁻(aq) + Cr²⁺(aq) ⇄ CrS(s)

The net ionic equation for the precipitation reaction is: Cr+ + 3S → CrS(s)

The net ionic equation for the precipitation reaction between potassium sulfide and chromium(II) nitrate can be written as:

Cr + 3S → CrS(s)

In this reaction, the chromium(II) ions (Cr) react with the sulfide ions (S) to form chromium(II) sulfide (CrS) which precipitates as a solid.

Learn more about Net ionic equation here:

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If the partial pressure of N2 in a scuba divers blood at the surface is 0.79 atm, what will the pressure be if he/she descends to a depth of 30 meters (4 atm) and stays there long enough to reach equilibrium (b)

Answers

Answer:

Explanation:

for every 3m that the internal pressure is lowered, it increases in an atmosphere approximately, so when the blood pressure of nitrogen decreases 30m, it will increase by approximately 10 atm, being enough there for the body to enter into equilibrium

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answers

Explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s

Answer:

(a)

$0.0342M$

(b)

$t_(1/2)=17.36s\nt_(1/2)=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$(1)/([NOBr])=kt +(1)/([NOBr]_0)\n(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\n$

$[NOBr]=(1)/(29.2/M)=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_(1/2)=(1)/(k[NOBr]_0)$

Therefore, for the given initial concentrations one obtains:

$t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\nt_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s$

Best regards.

A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.

Answers

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

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