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2 years ago

What is the molarity of a solution that contains 87.75g of NaCI in 500.ml of solution

Chemistry

1 answer:

Brut [27]2 years ago

6 0

Answer:

M = 3.0 mol/L.

Explanation:

We can calculate the molarity of a solution using the relation:

M = (mass x 1000) / (molar mass x V)

M is the molarity "number of moles of solute per 1.0 L of the solution.

mass is the mass of the solute (g) (m = 87.75 g of NaCl).

molar mass of NaCl = 58.44 g/mol.

V is the volume of the solution (ml) (V = 500.0 ml).

∴ M = (mass x 1000) / (molar mass x V) = (87.75 g x 1000) / (58.44 g/mol x 500.0 ml) = 3.0 mol/L.

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Give examples of how land resources are used in a city setting

poizon [28]

Answer:

Land resources such as gravel and bedrock are used to construct buildings, roads, and sidewalks. Land is also where city structures—such as buildings, roads, and sidewalks—are constructed. People use land resources such as bedrock and aggregate to construct buildings, roads, and sidewalks.

I hope this answer helps

5 0

2 years ago

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Which of the following are strongly hydrogen bonded in the liquid phase? A) nitrilesB) esters C) secondary amides D) acid chlori

-Dominant- [34]

Answer: Option (C) is the correct answer.

Explanation:

Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.

Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be strongly hydrogen bonded in the liquid phase of secondary amide.

Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.

Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.

6 0

3 years ago

Using the Law of Conservation of Matter, determine the number of grams of iron sulfide (FeS) that will be produced in this react

vazorg [7]

The Law of Conservation of Mass states that the mass of reactants entering a reaction must be equal to the mass of the products exiting it. In this case, we only have 2 reactants, Fe and S, and we only have 1 product, FeS. Therefore we expect the total mass of the Fe and S reactants to equal the mass of FeS. This gives us 112 g + 64 g = 176 g of FeS, which is choice D.

4 0

3 years ago

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Drag and drop each phrase below the type of weathering it describes.

iren2701 [21]

Answer:

Mechanical weathering

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C. abrasion

F. ice wedging

Chemical weathering

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Explanation:

5 0

2 years ago

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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

2 years ago