All categories

3 years ago

If a student needs to make exactly 2.5 liters of

1 answer:

4 0

260 mL

The concentration of the solution must go from 12.0 M, which is what you have for the stock solution, to 1.25 M.
The dilution factor will thus be
12/1.25= 9.6
This means that the volume of the stock solution used for the dilution was
2.5 L/9.6= 0.26 L
Expressed in milliliters and rounded to two sig figs, the answer will be 260 mL.

You might be interested in

Calculate the mass of a body<br>Whose volume is<br>Is 2cm3 and<br> density is 520cm3

MrRissso [65]

Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

6 0

2 years ago

What is the mole fraction of water in pure water

irinina [24]

In pure water, all of the molecules in the liquid are water molecules so the mole fraction is 1 (100 % H2O, 55 mol/L). In sea water, the concentration of water molecules in the solution is less than that of pure water so the vapor pressure of sea water is also lower.

5 0

3 years ago

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

muminat

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ?

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

$n = \dfrac{m}{MM}$

$5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}$

$m = 5.20*84.2$

$\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark$

_______________________

I Hope this helps, greetings ... Dexteright02! =)

4 0

3 years ago

Read 2 more answers

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Evgesh-ka [11]

Answer:

The answer to your question is P2 = 0.78 atm

Explanation:

Data

Temperature 1 = T1 = 263°K Temperature 2 = T2 = 298°K

Volume 1 = V1 = 24 L Volume 2 = V2 = 35 L

Pressure 1 = P1 = 1 Pressure 2 = P2 = ?

Process

1.- To solve this problem use the Combined gas law

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = P1V1T2 / T1V2

-Substitution

P2 = (1)(24)(298) / (263)(35)

-Simplification

P2 = 7152 / 9205

-Result

P2 = 0.777

or P2 = 0.78 atm

5 0

3 years ago

200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486

ololo11 [35]

The empirical formula of a given compound is C6H9ON5.

<u>Explanation</u>:

Step 1: Obtain the mass of each element present in grams

Element % = mass in g = m

Carbon = 83.884 grams, Hydrogen = 10.486 grams, Oxygen = 18.640 grams, Nitrogen = 86.99 grams.

Step 2: Determine the number of moles of each type of atom present

m/atomic mass = Molar amount (M)

Molar amount of carbon = (83.884 1 mol ) / 12 g = 6.99

Molar amount of hydrogen = (10.486 1 mol) / 1 g = 10.49

Molar amount of oxygen = (18.64 1 mol) / 16 g = 1.17

Molar amount of nitrogen = (86.99 1 mol) / 14 g = 6.21

Step 3: Divide the number of moles of each element by the smallest number of moles

M / least M value = Atomic Ratio (R)

Atomic radius of carbon = 6.99 / 1.17 = 5.9 = 6

Atomic radius of hydrogen = 10.49 / 1.17 = 8.9 = 9

Atomic radius of oxygen = 1.17 / 1.17 = 1

Atomic radius of nitrogen = 6.21 / 1.17 = 5

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

R * whole number = Empirical Formula

The empirical formula of a given compound is C6H9ON5.

7 0

3 years ago