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3 years ago

How many moles are in 3.45 g of Na2O?Round your answer to three significant

Chemistry

1 answer:

Neporo4naja [7]3 years ago

6 0

Answer: There are 0.056 moles present in 3.45 g of $Na_{2}O$ .

Explanation:

Given : Mass = 3.45 g

Moles is the mass of substance divided by its molar mass.

Hence, moles of $Na_{2}O$ (molar mass = 61.98 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{3.45 g}{61.98 g/mol}\\= 0.056 mol$

Thus, we can conclude that there are 0.056 moles present in 3.45 g of $Na_{2}O$ .

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What is the molarity of an aqueous solution that contains 78g of C6H12O6 dissolved in 2500 mL of solution?

dusya [7]

Answer:

$\boxed {\boxed {\sf molarity = 0.17 \ M \ C_6H_12O_6}}$

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

$molarity = \frac {moles}{liters}$

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.

1. Gram to Moles

We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.

C: 12.011 g/mol
H: 1.008 g/mol
O: 15.999 g/mol

Next, look at the formula and note the subscripts. This tells us the number of atoms in 1 molecule. We multiply the molar mass of each element by its subscript.

6(12.011)+12(1.008)+6(15.999)=180.156 g/mol

Use this number as a ratio.

$\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}$

Multiply by the given number of grams.

$78 \ g \ C_6H_12O_6 *\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}$

Flip the fraction and divide.

$78 \ g \ C_6H_12O_6 *\frac { 1 \ mol \ C_6H_12O_6}{180.156 \ g\ C_6H_12 O_6}$

$\frac { 78 \ mol \ C_6H_12O_6}{180.156 }= 0.432958102977 \ mol \ C_6H_12O_6$

2. Milliliters to Liters

There are 1000 milliliters in 1 liter.

$\frac {1 \ L }{ 1000 \ mL}$

Multiply by 2500 mL.

$2500 \ mL* \frac {1 \ L }{ 1000 \ mL}$

$2500 * \frac {1 \ L }{ 1000 }= 2.5 \ L$

3. Calculate Molarity

Finally, divide the moles by the liters.

$molarity = \frac {0.432958102977 \ mol \ C_6H_12O_6}{ 2.5 \ L}$

$molarity = 0.173183241191 \ mol \ C_6H_12O_6/L$

The original measurement has 2 significant figures, so our answer must have the same. That is the hundredth place and the 3 tells us to leave the 7.

$molarity \approx 0.17 \ mol \ C_6H_12O_6 /L$

1 mole per liter is also equal to 1 M.

$molarity = 0.17 \ M \ C_6H_12O_6$

3 0

3 years ago

Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re

lapo4ka [179]

a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K

when ΔG° is Gibbs free energy

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 = 298 K

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

= -7739 J = -7.7 KJ

b) then, to get E° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

= - 2.7 x 10^-2 V

8 0

3 years ago

Please help me!! thank you so much!!

Elis [28]

Answer:

13.75

Explanation:

gets halved every 8 days so halved twice

4 0

4 years ago

Compare the boiling points of 1-pentyne and 1-octyne.

Elan Coil [88]

1-pentyne consists of a carbon chain of 5 carbons one with a triple bond. 1-octyne is a carbon chain of 8 carbons with a triple bond at some point. It is known that the longer the carbon chain the higher the boiling point since more energy will be required to break the bonds between carbons. Based on this it is predicted that 1-octyne will have a higher boiling point than 1-pentyne.

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4 years ago

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Formula for "barium oxide"

Harman [31]

The answer is BaO
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4 years ago