Ask question

Ask question

All categories

3 years ago

14

A student placed 11.0 g of glucose (C6H12O6 ) in a volumetric flask, added enough water to dissolve the glucose by swirling, the

n carefully added additional water until the 100. mL mark on the neck of the flask was reached.
The flask was then shaken until the solution was uniform. A 55.0 mL sample of this glucose solution was diluted to 0.500 L .

How many grams of glucose are in 100. mL of the final solution?

1 answer:

Sedaia [141]3 years ago

4 0

Answer:

We have 12.1 grams of glucose in 100 mL of solution

Explanation:

Step 1: Data given

Mass of glucose = 11.0 grams

Volume = 100 mL

A 55.0 mL sample of this glucose solution was diluted to 0.500 L .

Step 2: Calculate concentration of glucose

Initially, amount of glucose available in 100 ml of the solution = 11.0 grams

Concentration of glucose = mass/ volume

Concentration=11.0 /0.100

= 110 g/L

We take away, 0.055 L of that solution, creating another solution with it but with 0.500 L of volume

Step 3: calculate concentration of the new volume

⇒ with C1 = Concentration of the first solution

⇒ with V1 = Volume of the first solution

⇒ with C2 = Concentration of the second solution

⇒ with V2 = Volume of the second solution

C1V1= C2V2

110 * 0.055= C2 * 0.500

= 12.1 g/L -= the concentration of the new solution

Step 4: Calculate the mass in 0.100 L

Concentration = mass/ volume

12.1 = mass/0.100

=1.21 grams

We have 12.1 grams of glucose in 100 mL of solution

You might be interested in

What is the name of the monomer that polymerizes to form polymethylmethacrylate

Cloud [144]

Answer:

polymethylmethacrylate bone cement is a polymer formed by mixing highly volatile liquid methylmethacrylate (MMA) monomer with an accelerator, polymethylmethacrylate powder. This bone cement is used during orthopedic surgery to implant prostheses for joint replacement

Explanation:

it was so easy thanks for the work form my brain

7 0

3 years ago

Approximately how many elements are there that combine chemically in a great number of ways to produce compounds? A. 25 B. 50 C

Arturiano [62]

100 as there’s approximately 100 discovered elements

8 0

3 years ago

Read 2 more answers

A sample of gold has a mass of 100 grams and a volume of 5cm^3, calculate the density by dividing the mass by volume

Dafna11 [192]

Answer:

20cm^2

Explanation:

Here, Density= Mass/ Volume

=100/5

= 20 cm^2

7 0

3 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

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

coldgirl [10]

Answer:

$\rm_{90}^{231}\text{Th}$

Explanation:

The unbalanced nuclear equation is

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} + X$

Let's write X as a nuclear symbol.

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} + _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

92 = 2 + Z , so Z = 92 - 2 = 90

And your nuclear equation becomes

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} +\, _{90}^{231}\text{X}$

Element 90 is thorium, so

$\rm X = _{90}^{231}\text{Th}$

7 0

3 years ago