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3 years ago

Calculate the molarity of 0.289 moles of FeC13 dissolved in 120 ml of solution

Chemistry

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

2.41 M

Explanation:

The molarity is the moles of FeCl3 over the liters of solution. Since you're given mL you need to change it to L which is 0.12 L. 0.289 divided by 0.12 is your answer

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Who discovered sodium?

Free_Kalibri [48]

Humphry Davy First discovered sodium

5 0

3 years ago

Read 2 more answers

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

3 years ago

Nitrosyl bromide decomposes according to the chemical equation below. 2NOBr(g) 2NO(g) + Br2(g) 1.00 atm of NOBr is sealed in a f

Andrew [12]

Given :

2NOBr(g) - -> 2NO(g) + Br2(g)

Initial pressure of NOBr , 1 atm .

At equilibrium, the partial pressure of NOBr is 0.82 atm.

To Find :

The equilibrium constant for the reaction .

Solution :

2NOBr(g) - -> 2NO(g) + Br2(g)

t=0 s 1 atm 0 0

$t=t_{eqb}$ 1( 1-2x) 2x x

So ,

$1-2x=0.82\\\\x=0.09$

At equilibrium :

$K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm$

Hence , this is the required solution .

3 0

3 years ago

What is the gram formula mass of Glycine , NH2CH2COOH

dem82 [27]

N=14
H= 1(5)= 5
C=12(2)=24
O=16(2)=32

= 75

6 0

3 years ago

What is the molarity of a nitric acid solution if 43.13 mL 0.1000 M KOH solution is needed to neutralize 30.00 mL of the acid so

Leno4ka [110]

Answer:

0.144M

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

HNO3 + KOH —> KNO3 + H20

From the equation,

nA = 1

nB = 1

From the question given, we obtained the following:

Ma =?

Va = 30.00mL

Mb = 0.1000M

Vb = 43.13 mL

MaVa / MbVb = nA/nB

Ma x 30 / 0.1 x 43.13 = 1

Cross multiply to express in linear form

Ma x 30 = 0.1 x 43.13

Divide both side by 30

Ma = (0.1 x 43.13) /30 = 0.144M

The molarity of the nitric acid is 0.144M

8 0

3 years ago