What number of moles of Pb (NO3)2 is consumed when 54 g of KNO3 is produced

How many moles of ba(oh)2 are present in 275 ml of 0.200 m ba(oh)2?

DENIUS [597]

You have to put your attention to the unit of concentration. It is expressed in terms of molarity, which is represented in M. It is the number of moles solute per liter solution. So, you simply have to multiply the molarity with the volume in liters.

Volume = 275 mL * 1 L/1000 mL = 0.275 L
Moles Ba(OH)₂ = (0.200 M)(0.275 L) = 0.055 mol

6 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

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Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago

DDT is currently banned worldwide. a. True b. False

Aleks [24]

B. False. It is banned worldwide for agricultural uses. The use of DDT is still authorized in small proportions in countries that need it, with support deployed for the alteration to safer and more effective alternatives.

5 0

3 years ago

Plz answer very easy question

Vinvika [58]

A: they all have protons
b:they have a different amount of neutrons
c:protons neutrons and electrons
d: the atom itself would change

7 0

3 years ago

If you pump 40.088g of octane (C8H18) into your gas tank, how mant molecules of octane are you pumping

Sonbull [250]

Answer:

2.11 × 10²³ molecules of octane

Explanation:

Given data:

Mass of octane = 40.088 g

Molecules of octane = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 40.088 g / 114.23 g/mol

Number of moles = 0.351 mol

Number of molecules:

one mole of octane = 6.022 × 10²³ molecules of octane

0.351 mole of octane = 0.351 × 6.022 × 10²³ molecules of octane

2.11 × 10²³ molecules of octane

6 0

3 years ago

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