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3 years ago

How many Moles of NAOH are present in 2.50L of 0.300 M NaOH?

Chemistry

1 answer:

maw [93]3 years ago

8 0

<h3>Answer:</h3>

0.75 moles NaOH

<h3>Explanation:</h3>

We are given;

Volume of NaOH solution = 2.5 Liters

Molarity of NaOH = 0.300 M

We are required to calculate the moles of NaOH

We need to establish the relationship between moles, molarity and volume of a solution.

That would be;

Concentration/molarity = Moles ÷ Volume

Therefore;

Moles = Concentration × Volume

Thus;

Moles of NaOH = 0.300 moles × 2.50 L

= 0.75 moles

Therefore, the number of moles of NaOH is 0.75 moles

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Which equation represents a redox reaction?

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Answer:

D. $2NaBr + Cl_2\rightarrow 2NaCl + Br_2$

Explanation:

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In this case, for the given set of chemical reactions, it is possible to infer that D. is a categorized as redox due to the following:

Since both chlorine and bromine remain as diatomic gases, their oxidation states in such a form is 0, but as anions with lithium cations they have a charge of - according to the following reaction and half-reactions:

$2NaBr + Cl_2\rightarrow 2NaCl + Br_2$

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3 years ago

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A sample of krypton gas in a container of volume 1.90 L exerts a pressure of 0.553 atm at 21 Celsius. How many moles of gas are

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Explanation:

pv = nRT

n= PV/RT

n =0.553×1.90/0.0821×294

n=1.0507/24.1374

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3 years ago

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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi

Gelneren [198K]

Answer : The normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of ethanol at $30^oC$ = 98.5 mmHg

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

$T_1$ = temperature of ethanol = $30^oC=273+30=303K$

$T_2$ = normal boiling point of ethanol = ?

$\Delta H_{vap}$ = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})$

$T_2=348.67K=348.67-273=75.67^oC$

Hence, the normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

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3 years ago

A simple circuit consists of a battery to provide power, wires to carry the ______, and load that uses the _________-for example

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A simple circuit consists of a battery to provide power, wires to carry the electrical power, and
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3 years ago

Calculate the ΔHrxn for the following

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Answer:

2 NO (g) → N2 (g) + O2 (g)

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2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)

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The ΔH for the reaction is -103.4 kJ

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3 years ago