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3 years ago

What is the molarity of a solution that contains 15.0g of naoh per 500.0 ml of solution

Chemistry

1 answer:

VARVARA [1.3K]3 years ago

8 0

Answer:

0.4 M

Explanation:

Molarity is defined as moles of solute, which in your case is sodium hydroxide,

NaOH

, divided by liters of solution.

molarity

moles of solute

liters of solution

Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,

Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to

determine how many moles of sodium hydroxide you have in that many grams

convert the volume of the solution from milliliters to liters

So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.

⋅

1 mole NaOH

40.0

0.175 moles NaOH

The volume of the solution in liters will be

500

⋅

1 L

1000

0.5 L

Therefore, the molarity of the solution will be

0.175 moles

0.5 L

0.35 M

Rounded to one sig fig, the answer will be

0.4 M

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Answer:

The atomic number of Selenium is 34. This means that Selenium possesses 34 electrons.

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7 0

3 years ago

Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g

Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

$\frac{Mass}{Volume}$

$Mass = \frac{percent \;of \;ethanol \;X \;density \;of \;ethanol \;+ \\ \;percent \;of \;water X \;density \;of \;water}{100}$

$0.926 = \frac{x X 0.7890 g/mL \;+ (100-x) X 0.9982 g/mL}{100}$

$x= 34.51$ %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

Learn more about the density here:

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2 years ago

Consider the molecules: scl2, f2, cs2, cf4, and brcl. select the choice where the bonds of the molecules are arranged in order o

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The polarity of a bond depends on the difference in electronegativities of the atoms involved in bonding. Greater this difference, greater is the polarity

1) In SCl2 we have the S-Cl bond

Electronegativity values: S = 2.5, Cl = 3

Difference = 3.0 -2.5 = 0.5

2) In F2 we have the F-F bond

Electronegativity value: F = 4.0

Difference = 4.0-4.0 = 0

3) In CS2 we have the C-S bond

Electronegativity value: C = 2.5, S = 2.5

Difference = 2.5 -2.5 = 0

4) In CF4 we have the C-F bond

Electronegativity values: C = 2.5, F = 4.0

Difference = 4.0 -2.5 = 1.5

5) In BrCl we have the Br-Cl bond

Electronegativity values: Br = 2.8, Cl = 3

Difference = 3.0 -2.8 = 0.2

Therefore, the increasing order of polarity is

F2 = CS2 < BrCl < SCl2 < CF4

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3 years ago

During the combustion of methane (CH4), one mole of methane gas combines with two moles of oxygen gas (O2) to produce one mole o

astraxan [27]

Answer:

CH4(g) +2O2____ CO2(g) +2H2O (l)

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3 years ago

The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 x 10-15 s-1 at 298 K and a rate constant of 8

laiz [17]

Answer:

The correct answer is 160.37 KJ/mol.

Explanation:

To find the activation energy in the given case, there is a need to use the Arrhenius equation, which is,

k = Ae^-Ea/RT

k1 = Ae^-Ea/RT1 and k2 = Ae^-Ea/RT2

k2/k1 = e^-Ea/R (1/T2-1/T1)

ln(k2/k1) = Ea/R (1/T1-1/T2)

The values of rate constant k1 and k2 are 3.61 * 10^-15 s^-1 and 8.66 * 10^-7 s^-1.

The temperatures T1 and T2 are 298 K and 425 K respectively.

Now by filling the values we get:

ln (8.66*10^-7/3.61*10^-15) = Ea/R (1/298-1/425)

19.29 = Ea/R * 0.001

Ea = 160.37 KJ/mol

3 0

3 years ago

What is the molarity of a solution that contains 15.0g of naoh per 500.0 ml of solution ​

What is the molarity of a solution that contains 15.0g of naoh per 500.0 ml of solution