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GenaCL600 [577]
4 years ago
14

KNO3(s) --> K+(aq) + NO3-(aq)This reaction was carried out in a Styrofoam insulated calorimeter and the following data were r

ecorded:Mass of solid KNO3 dissolved 10.1 gMass of aqueous solution (c = 4.18 J/gºC) 100. gT initial 30.0ºCT final 21.6ºCMolar mass of KNO3 101 g/molWhich of the following equations correctly shows the heat of solution (kJ/mol) for the dissolving of KNO3?KNO3(s) --> K+(aq) + NO3-(aq) + 3510 kJKNO3(s) --> K+(aq) + NO3-(aq) + 8.4 kJKNO3(s) + 35.1 kJ --> K+(aq) + NO3-(aq)KNO3(s) +3.51 kJ --> K+(aq) + NO3-(aq)
Chemistry
1 answer:
nataly862011 [7]4 years ago
4 0

Answer:

Equation correctly showing the heat of solution

KNO_3(s)+35.2 kJ \rightarrow K^+(aq) + NO_{3}^-(aq)

Explanation:

Mass of aqueous solution = m = 100 g

Specific heat of solution = c = 4.18 J/gºC

Change in temperature = \Delta T=T_f-T_i

ΔT = 21.6ºC - 30.0ºC = -8.4ºC

Heat lost by the solution = Q

Q = mc\Delta T

Q=100 g\times 4.18 J/g^oC\times (-8.4^oC)

Q = -3,511.2 J ≈ -3.51 kJ

Heat absorbed by potassium nitrate when solution in formed; Q'

Q' = -Q = 3.51 kJ

Moles of potassium nitrate , n= \frac{10.1 g}{101 g/mol}=0.1 mol

KNO_3(s)\rightarrow K^+(aq) + NO_{3}^-(aq)

The heat of solution =

\frac{Q'}{n}=\frac{3.5112 kJ}{0.1 mol}=35.1 kJ/mol

So, the equation correctly showing the heat of solution

KNO_3(s)+35.2 kJ \rightarrow K^+(aq) + NO_{3}^-(aq)

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How many grams of Br are in 335g of CaBr2
ASHA 777 [7]
The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
7 0
3 years ago
Use a volumetric flask to prepare 100.00 mL of 0.70M HCl solution. Do this by diluting either one of the stock hydrochloric acid
Orlov [11]

Answer:

- Take 3.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.

- Take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL

Explanation:

Hello,

In this case, given that the dilutions are preparedfrom 3.0-M and 6.0-M hydrochloric acid, we must proceed as follows:

- 3.0-M stock: when using this stock, the aliquot you must take is computed as shown below:

V_1=\frac{M_2V_2}{M_1}=\frac{100.00mL*0.70M}{3.0M}=23.3mL

It means that you must take 23.3 mL of 3.0-M hydrochloric acid and subsequently add 76.7 mL of water to complete the 100.00 mL.

- 6.0-M stock: when using this stock, the aliquot you must take is computed as shown below:

V_1=\frac{M_2V_2}{M_1}=\frac{100.00mL*0.70M}{6.0M}=11.7mL

It means that you must take 11.7mL of 6.0-M hydrochloric acid and subsequently add 88.3 mL of water to complete the 100.00 mL.

Regards.

3 0
3 years ago
The line spectrum of lithium has a red line at 670.8 nm. Calculate the energy of a photon with this wavelength.
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3x10^8 / (670.8 * 10^-9) =4.47x10^14 Hz 4.47x10^14 Hz multiplied by plank's constant = 2.9634x10^-19

2.96 x10-19 J
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3 years ago
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