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3 years ago

How many grams of sodium hydroxide are present in 247.0 mL of a 0.300 M NaOH solution?

Chemistry

1 answer:

mamaluj [8]3 years ago

5 0

The grams of NaOH that are present in 247.0 ml of a 0.300M NaOH solution is 2.964 grams

calculation

Step 1: find moles of NaOH

moles = molarity x volume in liters

volume in liters = 247.0 /1000 = 0.247 liters

molarity = 0.300 M = 0.300 mol/l

moles = 0.300 mol/l x 0.247 =0.0741 moles

step 2: find mass of NaOH

mass= moles x molar mass

The molar mass of NaOH = 23 +1 +16= 40 g/mol

mass=0.0741 moles x40 g/mol =2.964 grams

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Determine how many liters of hydrogen adjusted to STP there are in a 50.0 liter steel cylinder if the pressure inside is 100.0 a

Dvinal [7]

Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 100.0 atm

$P_2$ = final pressure of gas at STP = 1 atm

$V_1$ = initial volume of gas = 50.0 L

$V_2$ = final volume of gas at STP = ?

$T_1$ = initial temperature of gas = $27.0^oC=273+27.0=300K$

$T_2$ = final temperature of gas at STP = $0^oC=273+0=273K$

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

$\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}$

$V_2=4550L$

Therefore, the volume of hydrogen gas at STP is 4550 L.

3 0

3 years ago

A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.

Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Concentration of the weak acid (Ca): 0.187 M

pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

$Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}$

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2 years ago

This is due at 12am HELP!!

Lisa [10]

Answer:

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Explanation:

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4 0

3 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

Answer: The mass of HCl present in 500 mL of acid solution is 36.5 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0

2 years ago

How many grams of H,O are needed to produce 13 g of NaOH? 2 Na,O2 + 2 H20-4 NaOH + O2

fgiga [73]

Answer:

This is a typical stoichiometry question.To answer this question you want to get a relationship between

O and NaOH.

So you can get a relationship between the moles of

and moles of NaOH by the concept of stoichiometry.

O +

O ----------------> 2 NaOH.

According to above balanced equation we can have the stoichiometry relationship between

O and NaOH. as 1:2

It means 1 moles of

O is required to react with one mol of

O to produce 2 moles of NaOH.

in terms of mass 1 mole of

O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.

62 g of

O produces 80 g of NaOH.

1g of NaOH is produced from 62/80 g of

1.6 x

g of NaOH will require 62 x 1.6 x

g / 80 of

124g of

Explanation:

3 0

3 years ago