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pickupchik [31]
3 years ago
14

Briefly discuss interpretations of your observations and results. Include in your discussion, any conclusions drawn from the res

ults and any sources of error in the experiment. Be sure to discuss the reasons for your measured value of the specific heat of the metal being too high or too low.
Chemistry
1 answer:
VMariaS [17]3 years ago
4 0

Answer:

In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab

Explanation:

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A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet
stira [4]

Explanation:

The given data is as follows.

           Mass of apple sauce mixture = 454 kg

           Heat added (Q) = 121300 kJ

 Heat capacity (C_{p}) of apple sauce at 32.8^{o}C = 4.0177 kJ/kg^{o}C

So, Heat given by heat exchanger = heat taken by apple sauce

                            Q = mC_{p} \Delta T

or,                    Q = mC_{p} (T_{f} - T_{i})  

Putting the given values into the above formula as follows.

                     Q = mC_{p} (T_{f} - T_{i})  

              121300 kJ = 454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)

                      T_{f} = 76.5^{o}C

Thus, we can conclude that outlet temperature of the apple sauce is 76.5^{o}C.

3 0
3 years ago
Even though so much energy is required to form a metal cation with a 2+ charge, the alkaline earth metals form halides with gene
nasty-shy [4]

The ∆Hrxn of the reaction is -394.5kJ/mol.

<h3>What is alkaline earth metal? </h3>

The alkaline earth metals are those elements which correspond to group 2 of the modern periodic table.

All elements of this group forms a cation of +2 charge.

The other elements of this group are:

Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.

<h3>What is Halogen? </h3>

The Halogen elements are present in group 17 of the modern periodic table.

All elements of this groups forms anions of -1 charge.

The elements of this group are:

Fluorine, Chlorine, bromine, iodine, and astatine.

∆Hrxn = ∆H(bond broken) - ∆H(bond formed)

We have following data of bond energy in kJ/mol:

Mg—Mg = 738

Cl—Cl = -349

Mg—Cl = 783.5

Since, one mole of Mg react with one mole of Cl atom to form one mole of MgCl

∆Hrxn = 738-349-783.5

∆Hrxn = -394.5kJ/mol.

Thus, we concluded that the ∆Hrxn of the reaction is -394.5kJ/mol.

learn more about bond energy:

brainly.com/question/13526463

#SPJ4

3 0
1 year ago
Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1
tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0
2 years ago
Explain why the C―Ha bond is much more acidic than the C―Hb bond in pentan-2-one. Select the single best answer. 253 Ha is less
mafiozo [28]

Answer:

Ha is more acidic than Hb because loss of Ha forms a resonance-stabilized conjugate base.

Explanation:

The carbon atom that is next to the carbonyl group in pentan-2-one is known as the alpha carbon atom, this carbon atom bears the Ha, the alpha hydrogen atoms.

Ha is more acidic than Hb because, loss of Ha leads to the formation of a resonance stabilized enolate ion. This resonance stabilization of the ion formed makes loss of Ha an easier process than loss of Hb, hence the answer above.

3 0
3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
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