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agasfer [191]
2 years ago
15

heated blocks of iron used to be warm bed. A 1600g brick of iron heated to 155c would release how many joules of heat energy as

it cooled to 25c?
Chemistry
1 answer:
omeli [17]2 years ago
3 0
We are going to use this formula : 

Q = M*C*ΔT

when Q is the heat released by the material in J 

and M is the mass in gram = 1600 g 

C is the specific heat capacity of iron = 0.4494 J/g

and ΔT is the changing in temperature = 155 - 25 = 130 °C

So by substitution:

∴ Q = 1600 g * 0.4494 J/g * 130 °C

       = 93475 J

∴ The amount of heat released in Joule = 93475 J
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Solid aluminum metal and diatomic chlorine gas react spontaneously to form a solid product. Give the balanced chemical equation
ValentinkaMS [17]

When solid aluminum metal is reacted with diatomic chlorine gas, solid aluminum chloride is formed. This reaction is an example of synthesis or chemical combination in which two elements, aluminum and chlorine combine to form a new compound aluminum chloride.

Word equation: Aluminum (s)+ Chlorine (g)---> Aluminum chloride(s)

Molecular formula of the product formed is AlCl_{3}.

Therefore the balanced chemical equation representing the reaction of solid aluminum with gaseous dichlorine can be represented as,

2Al(s) + 3Cl_{2}(g)-->2AlCl_{3}(s)

8 0
2 years ago
A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition
lubasha [3.4K]

<u>Answer:</u> The pH of the solution after addition of KOH is 3.84

<u>Explanation:</u>

We are given:

pH of buffer = 3.75

pK_a of formic acid = 3.75

Using Henderson-Hasselbalch equation for formate buffer:

pH=pK_a+\log(\frac{[HCOO-]}{[HCOOH]})

Putting values in above equation, we get:

3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1

[HCOO-]=[HCOOH]

We are given:

Concentration of formate buffer = 0.100 M

[HCOO-]+[HCOOH]=0.1

[HCOO-]=[HCOOH]=0.05M

As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Putting values in above equation, we get:

1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol

The chemical reaction for formic acid and KOH follows the equation:

                  HCOOH+KOH\rightarrow HCOO^-+H_2O

Initial:       0.05    0.005               0.05

Final:         0.045          -                0.055          

Volume of solution = 500 + 5 = 505 mL = 0.505 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[HCOO^-]}{[HCOOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of formic acid = 3.75

[HCOO^-]=\frac{0.055}{0.505}

[HCOOH]=\frac{0.045}{0.505}

pH = ?

Putting values in above equation, we get:

pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84

Hence, the pH of the solution after addition of KOH is 3.84

3 0
2 years ago
When sample a of methane, ch4, decomposes, it produces 35.0 grams of c and 2.04 grams of h. when sample b of ch4 decomposes, it
Free_Kalibri [48]
Since both samples are pure CH4 (methane), the proportion of C to H that evolves from the decomposition should be equal. In equation form:
35.0 g C / 2.04 g H = 23.0 g C / x g H
Solving for x gives a value of x = 1.3406 g H
So 1.3406 grams of hydrogen will be produced from sample b.
7 0
2 years ago
In the reaction: pb + 2ag+ → pb2+ + 2ag, the oxidizing agent is
nika2105 [10]
With reference to present question, following things may be noted
1) Oxidation is process of lose of electron. Atom//ion undergoing oxidation is refereed as reducing agent
2) Reduction is process of gain of electron. Atom/ion undergoing reduction is refereed as oxidizing agent.

in present system, Ag+ gain an electron to reduced to Ag. Therefore, Ag+ is an oxidizing agent. 
6 0
2 years ago
Read 2 more answers
A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re
Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of \gamma.

\gamma=\frac{C_p}{C_v}

As, C_p=R+C_v

So, \gamma=\frac{R+C_v}{C_v}

Given :

C_v=20.8J/K\\\\R=8.314J/K

\gamma=\frac{8.314+20.8}{20.8}=1.4

Now we have to calculate the initial volume of gas.

Formula used :

P_1V_1=nRT_1

where,

P_1 = initial pressure of gas = 4.25 atm

V_1 = initial volume of gas = ?

T_1 = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)

V_1=5.80L

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

P_1V_1^{\gamma}=P_2V_2^{\gamma}

where,

P_1 = initial pressure of gas = 4.25 atm

P_2 = final pressure of gas = 2.50 atm

V_1 = initial volume of gas = 5.80 L

V_2 = final volume of gas = ?

\gamma = 1.4

Now put all the given values in above formula, we get:

(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}

V_2=8.47L

Now we have to calculate the final temperature of gas.

Formula used :

P_2V_2=nRT_2

where,

P_2 = final pressure of gas = 2.50 atm

V_2 = final volume of gas = 8.47 L

T_2 = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2

T_2=257.9K\approx 258K

Now we have to calculate the work done.

w=nC_v(T_2-T_1)

where,

w = work done = ?

n = number of moles of gas =1.0 mol

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = 258 K

C_v=20.8J/K

Now put all the given values in above formula, we get:

w=(1.0mol)\times (20.8J/K)\times (258-300)K

w=-873.6J

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0
2 years ago
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