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larisa86 [58]
3 years ago
15

if you weigh the lauric acid to the nearest 0.1g and you weigh the unknown solid to the nearest 0.0001g, can you calculate a mol

ar mass of your unknown solid to three significant figures? Show clearly your reasoning.
Chemistry
1 answer:
timama [110]3 years ago
3 0

Answer & Explanation:

The limiting significant figure in a calculation is our smallest figure. 1 g is one figure by using the significant figure rules. In the same manner we don't count zeros unless there is a non-zero number in front of it so .0001 g is still 1 figure.

So no we only know the unknown to one significant figure.

You might be interested in
La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácid
Tanya [424]

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácido salicílico se convirtieron en 55.45 g de aspirina. ¿Cuál fue el porcentaje de rendimiento?

<em>In English:</em>

Aspirin is prepared by reacting salicylic acid with excess ethanoic anhydride. In one experiment, 50.05 g of salicylic acid was converted to 55.45 g of aspirin. What was the yield percentage?

Answer:

el rendimiento porcentual para la cantidad dada de ácido salicílico es 84.99 %

<em>In English:</em>

<em>the percent yield for the given amount of salicylic acid is </em><em>84.99%</em>

<em></em>

Explanation:

La ecuación química equilibrada para la reacción se puede escribir como:

C₇H₆O₃ + C₄H₆O₃    →    C₉H₈O₄ + HC₂H₃O₂

Para la reacción mostrada arriba; El reactivo limitante de la reacción es el ácido salicílico. Ahora; calcular el porcentaje de rendimiento; se espera que primero determinemos el rendimiento teórico de la reacción.

Entonces; la fórmula para calcular el porcentaje de rendimiento: \mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }  

El rendimiento teórico se determina de la siguiente manera:

50.05 g * 1 mol / 138.21 g / mol de C₇H₆O₃ * 1 mol de C₉H₈O₄ / 1 mol de C₇H₆O₃ * 180.157 g / mol de C₉H₈O₄ = 65.24 g de C₉H₈O₄

Porcentaje de rendimiento \mathbf {= \frac{55.45 }{65.24 } *100 }

Porcentaje de rendimiento = 84.99%

Por lo tanto, el porcentaje de rendimiento para la cantidad dada de ácido salicílico es 84.99%

<em>In English:</em>

<em>The balanced chemical eqaution for the reaction can be written as:</em>

<em>C₇H₆O₃ + C₄H₆O₃    →    C₉H₈O₄ + HC₂H₃O₂</em>

<em>For the reaction shown above;  The limiting reactant from the reaction is  salicylic acid. Now; to calculate the percentage yield ; we are expected to first determine the theoretical yield of the reaction. </em>

<em>So; the formula for calculating the percentage yield </em>\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }<em>  </em>

<em />

<em>The theoretical yield is determined as follows:</em>

<em>50.05 g * 1 mol/ 138.21 g/mol of C₇H₆O₃ * 1 mol of C₉H₈O₄/ 1 mol of C₇H₆O₃ * 180.157 g/mol of C₉H₈O₄ = 65.24 g of C₉H₈O₄ is produced</em>

<em />

<em>Percentage yield </em>\mathbf {= \frac{55.45 }{65.24 } *100 }<em />

<em>Percentage yield = 84.99%</em>

<em />

<em>Thus, the percent yield for the given amount of salicylic acid is </em><em>84.99%</em>

7 0
3 years ago
Volume is an extensive physical property because it is dependent on the size of the sample.
BaLLatris [955]
Hi,

The statement is true, as the volume of a sample depends on its size.

I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
7 0
3 years ago
Read 2 more answers
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
Based on the passage, what is the structure of the product of the reaction between 8-hydroxyquinoline-5 sulfonate and HRP mcat r
dusya [7]

A because the outcome of this reaction exists a radical formed by the oxidation of an aromatic amine's or phenol's ring substituent. The hydroxyl group of a phenol serves as the ring substituent in this condition.

<h3>Which two enzyme types are required for the two-step process of converting cytosine to 5 hmC?</h3>
  • The methyl group exists moved to cytosine in the first step, and it exists then hydroxylated in the second stage.
  • Thus, a transferase and an oxidoreductase exist as the two groups of enzymes needed.

<h3>Which kind of interaction between proteins and the dextran column material is most likely to take place?</h3>
  • Hydrogen bonding because the glucose's OH would create an H-bond with any disclosed polar side chains on a protein surface.

<h3>Two out of the four proteins would adhere to a cation-exchange column at what buffer pH?</h3>
  • Only positively charged proteins can attach to a cation-exchange column, and this can only occur when the pH exists lower than the pI.
  • Proteins A and B would both be positively charged at pH 7.0.

To learn more about hydroxyquinoline refer to:

brainly.com/question/26102339

#SPJ4

4 0
2 years ago
How many moles in 3.69 x 10^30 molecules of carbon dioxide?
Sedaia [141]

0.612 \times 10^{7} \text { moles of } \mathrm{CO}_{2} are present in 3.69 \times 10^{30} \text { molecules of } \mathrm{CO}_{2}

<u>Explanation:</u>

It is known that each mole of an element is composed of avagadro's number of molecules. So if we need to determine, we need to divide the number of molecules with the avagadro's number.

So,

    1 mol of element =6.02 \times 10^{23} molecules of element

As here 3.69 \times 10^{30} molecules of carbon di oxide is given. So the moles in it will be

   No. of moles of carbon dioxide = \frac{3.69 \times 10^{30}}{6.02 \times 10^{23}}

    No. of moles = 0.612 \times 10^{7} moles of carbon dioxide.

Thus,

0.612 \times 10^{7} of carbon dioxide are present in 3.69 \times 10^{30} \text { molecules of } \mathrm{CO}_{2}.

4 0
3 years ago
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