If 2 compounds have the same empirical formula, they are the same (same properties, structure, etc.) *

My teacher never taught me how to do this helpppp!!!!!!!

djverab [1.8K]

Just search up how much the weight is equal to another, then multiply it, thats what i do lol sorry if im no help

6 0

3 years ago

A glucose solution that is prepared for a patient should have a concentration of 180 g/L. A nurse has 18 g of glucose. How many

Anuta_ua [19.1K]

You start by using proportions to find the number of liters of solution:

180 g of glucose / 1 liter of solution = 18 g of glucose / x liter of solution

=> x = 18 g of glucose * 1 liter of solution / 180 g of glucose = 0.1 liter of solution.

If you assume that the 18 grams of glucose does not apport volume to the solution but that the volume of the solution is the same volumen of water added (which is the best assumption you can do given that you do not know the how much the 18 g of glucose affect the volume of the solution) then you should add 0.1 liter of water.

Answer: 0.1 liter of water.

6 0

2 years ago

Read 2 more answers

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

Warm air and water both tend to rise while cooler air and water sink. When different parts of the oceans are heated unevenly, th

ale4655 [162]

Answer:

a

Explanation:

3 0

2 years ago

Help plz i need it my mom is mad at me for not knowing this

ohaa [14]

Lol ok ill help the answer is true and she SHOULDNT be mad at you

5 0

3 years ago