Help please i tried to do lab but i couldn’t figure it out

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

7 0

3 years ago

Fe + H2SO4 = Fe2(SO4)3 + H2

a_sh-v [17]

C (667) that’s the answer boiii

5 0

3 years ago

Help me please!! help would be appreciated

valina [46]

I don’t know if this is right but I thing It’s B

6 0

2 years ago

Read 2 more answers

What is the area of triangle ABC with vertices A(x¹,y¹), B(x²,y²)and C (x³,y³)??????????

AlexFokin [52]

Answer:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

$Area = 2\ units^2$

Explanation:

Given

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

Required

Determine the area

The area of a triangle is :

$Area = \frac{1}{2}|A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|$

By substituting values for the x and y coordinates of A, B and C;

We have:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

So:

For instance

$A = (0,3)$

$B= (2,1)$

$C = (2,3)$

The area is:

$Area = \frac{1}{2}|0(1-3) + 2(3-3) + 2(3-1)|$

$Area = \frac{1}{2}| 2*0 + 2*2|$

$Area = \frac{1}{2}| 0 + 4|$

$Area = \frac{1}{2}|4|$

$Area = \frac{1}{2} * 4$

$Area = 2\ units^2$

7 0

2 years ago

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

3 years ago