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PolarNik [594]
3 years ago
8

A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The speci

fic heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g
Chemistry
1 answer:
Molodets [167]3 years ago
6 0

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

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Chemical bonds release energy when they break apart, and take in energy when they form. How does this observation help explain t
soldier1979 [14.2K]
ENERGY WOULD BE RELALISED, MEANING BONDS ARE BEING BROKEN, SO IT IS AN EXOTHERMIC REACTION
6 0
2 years ago
Read 2 more answers
If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
3 0
3 years ago
Read 2 more answers
At a given temperature, 2.6 atm of h2 and 3.14 atm of cl2 are mixed and allowed to come to equilibrium. the equilibrium pressure
Keith_Richards [23]

The solution would be like this for this specific problem:

<span>Given:

H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>

 

<span>
pressure H2 = 2.6 - x 
pressure Cl2 = 3.14 - x 
<span>pressure HBr = 2x = 1.13

x = 1.13 / 2 = 0.565 

<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575 

Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>

 

= 1.2769 / (5.240125)

= 0.24367739319195629875241525726963

= 0.244

<span>Therefore, the Kp for the reaction at the given temperature is 0.244.

To add, </span>the hypothetical pressure of a gas if it alone occupied the whole volume of the original mixture at the same temperature is called the partial pressure or Kp.

7 0
2 years ago
I didn’t study for this lol if your answer correct I’ll mark ur answer brainliest
kotykmax [81]

Answer:

3

Explanation:

3 is the answer is did this a little while ago

4 0
3 years ago
Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w
jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, COCl_2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The value of K_c for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which p_{CO}=p_{Cl_2}=0.265atm and p_{COCl_2}=0.000atm ?

<u>Answer:</u> The equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of K_c\text{ and }K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p = Equilibrium constant in terms of partial pressure

K_c = Equilibrium constant in terms of concentration = 5.79

\Delta n_g = Difference between gaseous moles on product side and reactant side = n_{g,p}-n_{g,r}=1-2=-1

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

T = Temperature = 570 K

Putting values in above equation, we get:

K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

<u>Initial:</u>            0.265      0.265

<u>At eqllm:</u>        0.265-x    0.265-x        x

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}

Putting values in above equation, we get:

0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = (0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of Cl_2=(0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of COCl_2=x=0.008atm

Hence, the equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0
3 years ago
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