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lana66690 [7]
3 years ago
10

A student dissolves 14.9 g of ammonium chloride (NH4Cl in 250. g of water in a well-insulated open cup. She then observes the te

mperature of the water fall from 20.0 °C to 16.0 °C over the course of 5.2 minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction: NH,Cl(s) – NH (aq) + Cl (aq) You can make any reasonable assumptions about the physical properties of the solution. Be sure answers you calculate using measured data are rounded to 2 significant digits. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction. O exothermic Is this reaction exothermic, endothermic, or neither? endothermic x I ? neither If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case. kJ
Calculate the reaction enthalpy AH xn per mole of NHACI.
Chemistry
1 answer:
ololo11 [35]3 years ago
6 0

Answer:

Endothermic

4.433kJ

15.91 kJ / mol

Explanation:

In water, ammonium chloride dissolves thus:

NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq)

As the student observed the temperature decreased, the reaction is <em>endothermic </em>because absorbs heat from surroundigns when occurs.

It is possible to find the absorbed heat using coffee cup calorimeter equation:

Q = m×C×ΔT

<em>Where Q is heat, m is mass of solution (250g + 14.9g = 264.9g), C is specific heat of solution (It is possible to assume specific heat of pure water, 4.184J/g°C), and ΔT is change in temperature (20.0°C-16.0°C = 4.0°C)</em>

Replacing:

Q = 264.9g×4.184J/g°C×4.0°C

Q = 4433 J = <em>4.433kJ</em>

<em></em>

As enthalpy is the change in heat per mole of reaction, moles of ammonium chloride that reacted were:

14.9g NH₄Cl × (1mol / 53.491g) = <em>0.2786 moles</em>

As heat produced per 0.2786moles were 4.433kJ, heat per mole of ammonium chloride is:

4.433kJ / 0.2786mol = <em>15.91 kJ / mol</em>

<em />

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1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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3 years ago
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Answer:

B) microscopic

Explanation:

A scanning tunneling microscope allows imaging of microscopic particles.

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The isotopes K-37 and K-42 have the same
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Answer:

the same number of protons

Explanation:

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Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined
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Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

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3 years ago
A 3.31-g sample of lead nitrate, Pb(NO3)2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.53 L. T
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Answer:

0.486atm is the pressure of the cylinder

Explanation:

1 mole of Pb(NO₃)₂ descomposes in 4 moles of NO2 and 1 mole of O2. That is 1 mole descomposes in 5 moles of gas.

To find the pressure of the cylinder, we need to find moles of gas produced, and using general gas law we can determine the pressure of the gas:

<em>Moles Pb(NO₃)₂ and moles of gas:</em>

3.31g * (1mol / 331g) = 0.01 moles of Pb(NO₃)₂.

That means moles of gas produced is 0.05 moles.

<em>Pressure of the gas:</em>

Using PV = nRT

P = nRT/V

<em>Where P is pressure (Incognite)</em>

<em>V is volume (2.53L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature (300K)</em>

And n are moles of gas (0.05 moles)

P = 0.05mol*0.082atmL/molK*300K / 2.53L

P = 0.486atm is the pressure of the cylinder

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