Ask question

Ask question

All categories

love history [14]

2 years ago

8

"You have a solution of glucose in water that has a concentration of 2.50 M and a volume of 0.442 liters. You dilute this soluti

on with water to make a total volume of 1.50 liters. What is the final concentration of this solution?"

1 answer:

taurus [48]2 years ago

7 0

Answer:

0.737M

Explanation:

C1V1=C2V2

2.50 x 0.442 = C x 1.5

1.105 = C x 1.5

C = 1.105/1.5

C = 0.737M

You might be interested in

Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh

Ann [662]

Answer:

See explanation and image attached

Explanation:

The Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi. R is an alkyl or aryl group.

They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.

In this particular instance, we intend to synthesize propylcyclohexane. The structure of the lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.

7 0

2 years ago

HELP PLS THANK YA!!!

eduard

Answer:

there is the answers. hope this helps

5 0

2 years ago

Does anyone know how to write out the formula for these?? not just the answer x

Tju [1.3M]

Idk how to write the formula I had this same question

5 0

2 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

2 years ago

How many liters of space will 7.80 moles of methane gas (CH4) occupy at STP

Alchen [17]

Answer:

V CH4(g) = 190.6 L

Explanation:

assuming ideal gas:

PV = RTn

∴ STP: T =298 K and P = 1 atm

∴ R = 0.082 atm.L/K.mol

∴ moles (n) = 7.80 mol CH4(g)

∴ Volume CH4(g) = ?

⇒ V = RTn/P

⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)

⇒ V CH4(g) = 190.6 L

4 0

2 years ago