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2 years ago

How many moles of naf are in 34.2 grams of a 45.5% by mass naf solution?

1 answer:

8 0

<span>% by mass = mass solute x 100 / mass solution

45.5 % = mass solute NaF x 100 / 34.2

mass solute NaF = 34.2 x 45.5 /100=15.6 g

molae solute NaF = 15.6 g ( 1 mol / 41.9887 g)= 0.372</span>

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3 years ago

Yet a third pair of compounds of manganese and oxygen is 50.48% and 36.81% oxygen respectively. In what small whole number ratio

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Answer:

The number ratio is 4:7

Explanation:

Step 1: Data given

Compound 1 has 50.48 % oxygen

Compound 2 has 36.81 % oxygen

Molar mass oxygen = 16 g/mol

Molar mass manganese = 54.94 g/mol

Step 2: Calculate % manganes

Compound 1: 100 - 50.48 = 49.52 %

Compound 2: 100 - 36.81 = 63.19 %

Step 3: Calculate mass

Suppose mass of compounds = 100 grams

Compound 1:

50.48 % O = 50.48 grams

49.52 % Mn = 49.52 grams

Compound 2:

36.81 % O = 36.81 grams

63.19 % Mn = 63.19 grams

Step 4: Calculate moles

Compound 1

Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles

Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles

Compound 2

Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles

Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles

Step 5: calculate mol ratio

We will divide by the smallest amount of moles

Compound 1

O: 3.155/0.9013 = 3.5

Mn: 0.9013 / 0.9013 = 1

Mn2O7

Compound 2

O: 2.301 / 1.150 = 2

Mn: 1.150 / 1.150 = 1

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The number ratio is 2:3.5 or 4:7

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3 years ago

For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con

Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

$A\rightarrow B$

Expt [A] M [B] M Rate [M/s]

1 3.40 4.16 1.82*10^-4

2 4.59 4.16 3.32*10^-4

3. 3.40 5.46 1.82*10^-4

$Rate = k[A]^{x}[B]^{y}$

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

$\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2$

To find y:

Divide rate of expt 3 by expt 1

$\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0$

Therefore: x = 2, y = 0

$Rate = k[A]^{2}[B]^{0}$

To find k

Use rate for expt 1:

$k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1$

6 0

2 years ago