How much heat, in joules, must be added to 17 grams water (specific heat = 4.184

Question

How much heat, in joules, must be added to 17 grams water (specific heat = 4.184 J/g degreeC) in order to raise the temperature from 27 to 74 What is the molality of a solution that is 13.0% methanol (CH_3O) by mass if its density is 0.998 g/mL In a 1.0L vessel, at 698 K,0180 M H_2 (g) and 0.180 MI_2 (g) are combined. What is the concentration of HI (g) at equilibrium Kc = 54.3 at 698 K H_2 (g) + I_2 (g) left arrow right arrow 2 HI (g) The value of Kc is determined to be 1.62 for the following equation at -35 degrees C. NO, (g) N0_2 (g) left arrow right arrowN_2 O_4 (g) What is the value of Kp tor this reaction The value of Kc is determined to be 0.407 for the following equation at a particular temperature. 2NO_2 (g) left arrow right arrow N_2O_4 (g) What is the value of Kc for the following reaction 2N_2O_4 left arrow right arrow 4 NO_2 (g) How many neutrons does the reactant (X) contain X right arrow^204 Pt + 2 alpha The first order rate constant for a reaction is 1.60 * 10^-4 sec_1. What is the half-life of the reaction in seconds What is the frequency factor, A for a reaction whose energy of activation is 3.20 * 10^3 J/mol at 414 K The rate constant for this reaction was determined to be 9.84 * 10^-4 seconds^-1

Explanation / Answer

Solution :-

Q10) 17 g water , S=4.184 J per g C, T1 = 27 C , T2 = 74 C, q= ?

Formula

q= m*s*delta t

q= 17 g * 4.184 J pe g C * (74 C -27 C)

q= 3343 J

so the amount of heat needed = 3343 J

Q11) 13% methanol , density = 0.998 g/ml

Lets assume we have 1 L solution

1 L= 1000 ml

Therefore mass of solution = 1000 ml * 0.998 g per ml = 998 g

Now lets calculate the mass of methanol

998 g * 13 % methanol / 100 % = 129.74 g methanol

Now lets calculate the moles of methanol

Moles = mass / molar mass

Moles of methanol = 129.74 g / 32.04 g per mol = 4.05 mol methanol

Molarity = moles / volume in liter

Molarity of methanol = 4.05 mol / 1 L

                                      = 4.05 M

So the molarity of methanol = 4.05 M

Q12) H2 = 0.180 mol and I2 =0.180 mol

Volume = 1 L

So the molarity of the H2 = 0.180 M and I2 = 0.180 M

Kc= 54.3

   H2     +   I2 -------- > 2HI

0.180        0.180             0

-x                -x                  +2x

0.180-x     0.180-x           2x

Kc = [HI]^2/[H2][I2]

54.3 = [2x]^2/[0.180-x][0.180-x]

54.3 *[0.180-x]^2 = 4x^2

Solving for x we get

x= 0.142

therefore now lets calculate the HI concentration at equilibrium

[HI] = 2x = 2*0.142 M = 0.284 M

Q13) Kc =1.62

T = -35 C +273 = 238 K

Kp = ?

2NO2 ------ > N2O4

Delta n = sum of coefficient of product – sum of coefficient of reactant

             = 1-2

            = -1

Formula to calculate the Kp is as follows

Kp = Kc*(RT)^delta n

Kp = 1.62 *(0.08206 L atm per mol K * 238 K)^-1

Kp = 0.0829

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